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MCQ Questions for Class 10 Maths Application of Trigonometry with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 9 Application of Trigonometry Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Application of Trigonometry MCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 9 Application of Trigonometry

MCQ On Application Of Trigonometry Class 10 Question 1. The shadow of a tower is equal to its height at 10-45 a.m. The sun’s altitude is (a) 30° (b) 45° (c) 60° (d) 90°

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 9

Answer: a Explaination: Reason: In rt. ∆EBC, cos 60° = \(\frac{BC}{CE}\) ⇒ \(\frac{1}{2}\) = \(\frac{6}{CE}\) ⇒ CE = 12 cm

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 2

Answer: d Explaination: Reason: In rt. ∆ABC, cos C = \(\frac{BC}{AB}\) = \(\frac{7}{14}\) = \(\frac{1}{2}\) ⇒ cos C = cos 60° ∴ C = 60°

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 3

Answer: c Explaination: Reason: In rt ∆ADB, AB² = AD² + BD² = (4)² + (3)² = 16 + 9 = 25 ∴ AB = √25 = 5 ∴ In rt ∆ABC, tan θ \(\frac{AB}{BC}\) = \(\frac{5}{12}\)

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 7

Answer: d Explaination: Reason: Since ABCD is a rectangle ∴ BC = AD = 8 cm and B = 90° In rt ∆CBE, cos 60° = \(\frac{CE}{BC}\) ⇒ \(\frac{1}{2}\) = \(\frac{CE}{8}\) ∴ CE = \(\frac{8}{2}\) = 4 cm

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 8

Answer: c Explaination: Reason: Since ABCD is a || gm ∴ AD = BC = 4√3 In rt ∆APD, sin 60° = \(\frac{AP}{AD}\) ⇒ \(\frac{\sqrt{3}}{2}=\frac{\mathrm{AP}}{4 \sqrt{3}}\) ⇒ 2AP = 4 × 3 = 12 ∴ AP = 6 cm

MCQ On Trigonometry For Class 10 With Solutions Question 10. When the length of shadow of a vertical pole is equal to √3 times of its height, the angle of elevation of the Sun’s altitude is (a) 30° (b) 45° (c) 60° (d) 15°

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 13

MCQ Trigonometry Class 10 Question 11. The angle of elevation of top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. The length of the tower is (a) √3 m (b) 2√3 m (c) 5√3m (d) 10√3 m

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 14

MCQ Questions On Trigonometry For Class 10 Question 12. A plane is observed to be approaching the airport. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 60°. The height above the ground of the plane is (a) 6√3 m (b) 4√3 m (c) 3√3 m (d) 2√3 m

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 15

Trigonometry Multiple Choice Questions And Answers Pdf Question 13. The upper part of a tree is broken by the wind and makes an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 5 m. The height of the tree is (a) 10√33 m (b) 5√33 m (c) √3 m (d) √3/5 m

MCQ Questions for Class 10 Maths Application of Trigonometry with Answers 16

MCQ On Trigonometry 10th Question 14. The angles of elevation of the top of a rock from the top and foot of 100 m high tower are respectively 30° and 45°. The height of the rock is (a) 50 m (b) 150 m (c) 5o√3m (d) 50(3 + √3)

Trigonometry MCQs Question 15. The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, the length of the wire is (a) 6 m (b) 10 m (c) 12 m (d) 20 m

We hope the given MCQ Questions for Class 10 Maths Application of Trigonometry with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 9 Application of Trigonometry Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.

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MCQ Questions for Class 10 Introduction to Trigonometry with Answers

Students can refer to the following Trigonometry Class 10 MCQ Question with Answers provided below based on the latest curriculum and examination pattern issued by CBSE and NCERT. Our teachers have provided here a collection of multiple choice questions for Trigonometry Class 10 covering all topics in your textbook so that students can assess themselves on all important topics and thoroughly prepare for their exams

Trigonometry Class 10 MCQ Question with Answers

We have provided below chapter-wise Trigonometry Class 10 MCQ Question with answers which will help the students to go through the entire syllabus and practice multiple choice questions provided here with solutions. As Trigonometry MCQs in Class 10 pdf download can be really scoring for students, you should go thorough all problems and MCQ Questions for Class 10 Maths provided below so that you are able to get more marks in your exams.

Question. What is the value of sin 2 5 o + sin 2 10 o + sin 2 80 o + sin 2 85 o ? (a) 0 (b) 1 (c) 2 (d) 3

Question. If tan 26 o tan19 o / x(1 tan 26 o tan19 o ) = cos60 o , what is the value of x ? (a) 1 (b) 2 (c) 2 (d) 3

Question. If 4tanΘ = 1, find the value of 4sinΘ – 2cosΘ / 4sinΘ + 3cosΘ (a) 1/2 (b) 1/6 (c) 2/3 (d) 1/3

Question. Find the value of 5cos α – 4 / 3 + 5sin α / 3 – 5sin α / 4 + 5cos α (a) – 1 (b) 5 (c) 1 (d) 0

Question. If sin A + sin2 A = 1 , find the value of the expression (cos2 A + cos4 A) . (a) 1 (b) 1/2 (c) 2 (d) 3

Question. Find the value of sin 15 o  . (a) √3+1/√2 (b) √3 -1/√2 (c) √3+1/2√2 (d) √3 -1/2√2

Question. Find the value sin 2 30 o cos 2 45 o + 4tan 2 30 o + 1/2 sin 2 90o 2cos 2 90 o + 1/24 (a) 1 (b) 2 (c) √2 (d) √3

Question. An electric pole is 10√3 m high and its shadow is 10 m in length, then the angle of elevation of the sun is  (a) 45° (b) 15° (c) 30° (d) 60°

Question. A kite is flying at a height of 90 m above the groun(d) The string attached to the kite is temporarily tied to a point on the groun(d) The inclination of the string with the ground is 60° . The length of the string, assuming that there is no slack in the string is      (a) 90√3 m (b) 60√3 m (c) 90 m (d) 45 m

Question. If the angle of depression of a car from a 100 m high tower is 45°, then the distance of the car from the tower is    (a) 100 m (b) 200 m (c) 100√3m (d) 100√2m

Question. A kite is flying at a height of 200 m above the groun(d) The string attached to the kite is temporarily tied to a point on the groun(d) The inclination of the string with the ground is 45°. The length of the string, assuming that there is no slack in the string is    (a) 100 m (b) 200 m (c) 200√2 m (d) 100√2 m

Question. The _____________ is the line drawn from the eye of an observer to the point in the object viewed by the observer.      (a) Horizontal line (b) line of sight (c) None of these (d) Vertical line

Question. A pole 10 m high cast a shadow 10 m long on the ground, then the sun’s elevation is      (a) 60° (b) 15° (c) 45° (d) 30°

Question. Find sin 3 φ + cos 3  φ/sin φ + cos φ . (a) 1 + sin – cos φ (b) 1- sinc φ cos φ (c) 1- sin φ tan φ (d) 1

Question. lf sinθ + cosθ/sinθ – cosθ = √3 + 1/√3 – 1 find the acute angleθ . (a) 90° (b) 45° (c) 30° (d) 60°

Question. What is the value of θ tor which tanθ = cotθ ? (a) 60° (b) 45° (c) 90° (d) 0°

Question. The figure shows an isosceles triangle ABC. Find the length of the perpendicular from A to BC.

MCQ Questions for Class 10 Introduction to Trigonometry with Answers

(a) 5.45 cm (b) 4.55 cm (c) 5.6 cm (d) 4.54 cm

Question. What is the value of sinθ + cos30° – tan45°+ cosec 60° + cot90° ? (a) 7√3 – 6/6 (b) 6 + 7√3/6 (c) 0 (d) 2

Question. If sin θ = 1/2 , what are the respective possible values of 9 between 0 and 2π ? (a) 210° and 300° (b) 240° and 330° (c) 240° and 300° (d) 210° and 330°

Question. In terms of radians. what is the equivalent of 45° ? (a) 25π (b) 0.25π (c) 180°π /45° (d) 45°/π

Question. ΔABC is right angled at A. If AC = 8 cm and AB = 6 cm, what is the value of cosec B? (a) 5/4 (b) 3/4 (c) 4/3 (d) 4/5

Question. ΔABC is right angled at A .lf BC = √2 and AB = AC =1, what is the measure of ∠B ? (a) 60° (b) 45° (c) 30° (d) 90°

Question. Find the value of 5cos α – 4/3 – 5 sin α – 3 + 5 sin α/4 + 5 cos α . (a) – 1 (b) 5 (c) 1 (d) 0

Question. Two men are on opposite sides of a tower. They observe the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 100m, then the distance between them is    (a) 100(√3−1) m (b) 100(√3+1) m (c) 100(1−√3) m (d) none of these

Question. The ___________ is the angle between the horizontal and the line of sight to an object when the object is below the horizontal level.      (a) angle of projection (b) angle of elevation (c) None of these (d) angle of depression

Question. The ratio between the height and the length of the shadow of a pole is √3: 1, then the sun’s altitude is    (a) 45° (b) 30° (c) 75° (d) 60°

Question. The angle of elevation of the top of a tower from a point on the ground and at a distance of 30°m from its foot is . The height of the tower is    (a) 30√3m (b) 10 m (c) 10√3m (d) 30 m

Question. The ___________ of an object is the angle formed by the line of sight with the horizontal when the object is above the horizontal level.    (a) angle of projection (b) angle of depression (c) angle of elevation (d) none of these

Question. From a point on the ground which is 15m away from the foot of a tower, the angle of elevation is found to be 60° . The height of the tower is    (a) 15√3 m (b) 20√m (c) 10√m (d) 10 m

Question. From a point P on the level ground, the angle of elevation of the top of a tower is . If the tower is 100m high, the distance between P and the foot of the tower is    (a) 300√3 m (b) 150√3 m (c) 200√3 m (d) 100√3 m

Question. If the height of the tower is √3 times of the length of its shadow, then the angle of elevation of the sun is  (a) 15° (b) 30° (c) 50° (d) 45°

Question. A ramp for disabled people in a hospital must slope at not more than 30°. If the height of the ramp has to be 1 m, then the length of the ramp be    (a) 3 m (b) 1 m (c) 2 m (d) √3 m

Question. If the shadow of a boy ‘x’ metres high is 1.6m and the angle of elevation of the sun is then the value of ‘x’ is 45°  (a) 0.8 m (b) 1.6 m (c) 3.2 m (d) 2 m

Question. If the angle of depression of an object from a 75m high tower is 30° , then the distance of the object from the tower is    (a) 25√3 m (b) 50√3 m (c) 100√3 m (d) 75√3 m

Question. In a right triangle ABC, If AC = BC and then find its value      (a) 50√3  (b) 150 m (c) None of these (d) 100√3 m

Question. The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50m high, then the height of the hill is      (a) 50√3 m (b) 150m (c) 150√3 m (d) 100√3 m

Question. The angles of elevation of the top of a tower from two points on the ground at distances 8 m and 18 m from the base of the tower and in the same straight line with it are complementary. The height of the tower is      (a) 12 m    (b) 18 m (c) 8 m (d) 16 m

Question. If cos9α = sinα a and 9α < 90 o , what is the value of tan 5 a? (a) 1/√3 (b) √3 (c) 1 (d) 1/2

Question. ABCD is a trapezium in which AB = 8 cm, AD = 4 cm and CD = 3 cm. 

MCQ Questions For Class 10 Introduction to Trigonometry

What is the length of BC to the nearest whole number? (a) 5 cm (b) √41cm (c) 8 cm (d) 7 cm

Question. Find the value of cos1 o cos2 o cos3 o …..cos89 o cos90 o (a) 1 (b) 1/2 (c) 1/√2 (d) 0

Question. Find the value of 2/3 (cos 4 30 o – sin 4 45 o ) – 3 (sin 2 60 o – sec 2 45 o ) + 1/4 cot 2 30 o (a) 15/4 (b) 3/4 (c) 2(65/4) (d) 4(17/24)

Question. If 8 tan A =15, find the value of sinA – cosA / sinA + cosA (a) 7/23 (b) 11/23 (c) 13/23 (d) 17/23

Question. If 4sinΘ = 3cosΘ ,find sec 2 Θ / 4(1 – tan 2 Θ) (a) 25/16 (b) 25/28 (c) 1/4 (d) 5/6

Question. If ΔABC is right angled at C, find the value of cos(A + B) . (a) 0 (b) 1 (c) 1/2 (d) √3/2

Question. If the shadow of a tower is 30 m long when the sun’s elevation is . The length of the shadow, when the sun’s elevation is      (a) 10 m (b) 30 m (c) 10√3 m (d) 20 m

Question. Two men are on opposite sides of a tower. They observe the angles of elevation of the top of the tower as 60° and 45° respectively. If the height of the tower is 60m, then the distance between them is    (a) 20(√3−√3)m (b) 20(√3−√3)m (c) None of these (d) 20(√3+√3)m

Question. A river is 60 m wide. A tree of unknown height is on one bank. The angle of elevation of the top of the tree from the point exactly opposite to the foot of the tree, on the other bank, is 30°. The height of the tree is    (a) 30√3 m (b) 10√3 m (c) 20√3 m (d) 60√3 m

Question. A plane is observed to be approaching the airport. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 30° there at. Its height above the ground is    (a) 10 km (b) 12 km (c) 6 km (d) none of these

Question. If the altitude of the sun is 60° , the height of a tower which casts a shadow of length 90 m is    (a) 60 m (b) 90√3 m (c) 90 m (d) 60√3 m

Question. What is the angle between the hour and minute hands of a clock at 02 :15 hours? (a) 15 o (b) 7 (1 o /2) (c) 22(1 o /2) (d) 30 o

Question. What will be the change in the value of cosq if we decrease the value of θ? 

  It will increase

Question. What is the relation between sin θ ,cos θ and cotθ? 

cotθ= cosθ /sinθ

Question.If we increase the lengths of the sides of a right triangle keeping the angle between them same, then the values of the trigonometric ratios will also increase. State True or False. 

 False

Question. Does the value of tanq increase or decrease as we increase the value of θ? Give reason.        

Increase because as we increase q, the side opposite to right angle will increase and the ratio of tanq will also increase.

Question. What is the relation between tanq and secθ? 

  1 +tan 2 θ = sec 2 θ

Question. The value of tan A is always less than 1. State True or False. 

Question. Can the value of cos θ be 5/4 for some angle θ? 

Question. What is tan(90°-A) equal to? 

Question. What do we call the side opposite to the right angle in a right triangle?   

Question.  What is the reciprocal of sec A? 

Question. Is tan A the reciprocal of cot A?   

Question. Define an identity. 

An equation which holds true for all values of the variable.

Question. What is the maximum possible value for sine of any angle?   

Question. What is the value of sine of 0°?   

Question. What is 1+tan2θ? 

Question. What is the value of cosec 2θ -cot 2θ? 

Question. Name the side adjacent to angle A if DABC is a triangle right-angled at B.     

Question. Can the value of secant of an angle be greater than 1?        

Question If 2sin 2 θ -cos 2  θ = 2, then find the value of θ.

Question If cot θ = 7/8 check whether 1-tan 2 θ/1+tan 2 θ = cos 2 θ -sin 2 θ or not.

LHS = RHS = -15/113

Question.  Write True or False. Question In ΔABC, if ∠A +∠C = 90°, then sin A = cosC     

Question cot 60° = tan (90° – 30°)   

Question The value of 2sinθ can be a+1/a, where a is a positive number and a ≠ 1.

QuestionsecA =1/cosA = 1 , for an acute θ   

Question sin60° = 2sin30°   

Question tan θ increases faster than sinθ as θ increases. 

Question If cot θ = 7/8 check whether 1-tan 2 θ/1+tan 2 θ = cos 2 θ -sin 2 θ or not.

Question sinθ +cosθ =1       

Question cos75° = cos60° +cos15°   

Question The value of the expression (sin70° -cos70°) is negative.   

Question If cosA+cos 2 A =1, then sin 2 A+sin 4  A=1.    

Question If tan A = 3/4 , then cos A = 4/5   

Question tan 2  A = sec  2  A -1 (v) sin 2  56° +cos 2  34° =1       

Question cosec 50° = sec 40°       

Question cos A = cos × A     

Question cosθ = 7/6 for some angleθ   

Question. Fill in the blanks.

Question sin A _____________ when A increases from 0° to 90°.

Question cos A _____________ when A increases from 0° to 90°.

Question cot 90° is ___________. 

Question If cosθ =1, then θ = ___________. 

Question sin58°/cos32° = _____________.

Question  3tan 2 45° = ___________.   

Question  5 cos 0° + sin 90° = ___________.

Question  tan 0° = ___________. 

Question  2sin 2 45° = ___________.   

Question cos0° × cos10° × cos30° × cos80° × cos90° = _____________ .

Question The word ‘Trigonometry’ is derived from the Greek words _____________, _____________ and _____________.

Tri, gon, metron

Our teachers have developed really good  Multiple Choice Questions  covering all important topics in each chapter which are expected to come in upcoming tests and exams, as MCQs are coming in all exams now therefore practice them carefully to get full understanding of topics and get good marks. Download the latest questions with multiple choice answers for Class 10 Trigonometry in pdf or read online for free.

The above  NCERT based Trigonometry Class 10 MCQ Question have been designed by our teachers in such a way that it will help you a lot to gain an understanding of each topic. These CBSE NCERT Class 10 Trigonometry Multiple Choice Questions have been developed and are available free for benefit of Class 10 students.

Advantages of  Trigonometry Class 10 MCQ Question with Answers

a) MCQs will help the kids to strengthen concepts and improve marks in tests and exams.

b) Multiple Choice Questions for Trigonometry Class 10 have proven to further enhance the understanding and question-solving skills.

c) Regular reading topic wise questions with choices will for sure develop very good hold over each chapter which will help in exam preparations.

d) It will be easy to revise all Trigonometry chapters and faster revisions prior to class tests and exams.

Free Printable  MCQs in PDF of CBSE Class 10 Trigonometry is designed by our school teachers and provide the best study material as per CBSE NCERT standards.

MCQ Questions for Class 10 Trigonometry with Answers

You can easily get MCQs for Trigonometry from  https://www.cbsencertsolutions.com

The MCQs for Class 10 Trigonometry with Answers have been developed based on the current  NCERT textbook  issued by CBSE.

MCQs cover the topics of all chapters given in NCERT Book for Class 10 Trigonometry .

Yes – These Multiple Choice Questions for Class 10 Trigonometry with Answers are free to print and use them later.

No – All MCQs for Trigonometry are free to read for all students.

Just scroll and read the free MCQs.

Yes – you can download free MCQs in PDF for Trigonometry in standard MCQs format with Answers.

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MCQ Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Answers

Students can access the NCERT MCQ Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 10 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Introduction to Trigonometry Class 10 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 10 Maths Chapter 8 Introduction to Trigonometry Objective Questions.

Introduction to Trigonometry Class 10 MCQs Questions with Answers

Students are advised to solve the Introduction to Trigonometry Multiple Choice Questions of Class 10 Maths to know different concepts. Practicing the MCQ Questions on Introduction to Trigonometry Class 10 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Introduction to Trigonometry Class 10 with answers provided with detailed solutions by looking below.

Question 1. If cos (α + β) = 0, then sin (α – β) can be reduced to (a) cos β (b) cos 2β (c) sin α (d) sin 2α

Answer: (b) cos 2β

Question 2. If cos (40° + A) = sin 30°, the value of A is:? (a) 60° (b) 20° (c) 40° (d) 30°

Answer: (b) 20°

Question 3. If sin x + cosec x = 2, then sin 19 x + cosec 20 x = (a) 2 19 (b) 2 20 (c) 2 (d) 2 39

Answer: (c) 2

Question 4. If cos 9a = sin a and 9a < 90°, then the value of tan 5a is (a) \(\frac { 1 }{ \sqrt { 3 } }\) (b) √3 (c) 1 (d) 0

Answer: (c) 1

Question 5. 7 sin 2 θ + 3 cos 2 θ = 4 then : (a) tan θ = \(\frac { 1 }{ \sqrt { 2 } }\) (b) tan θ = \(\frac { 1 }{ 2 }\) (c) tan θ = \(\frac { 1 }{ 3 }\) (d) tan θ = \(\frac { 1 }{ \sqrt { 3 } }\)

Answer: (d) tan θ = \(\frac { 1 }{ \sqrt { 3 } }\)

Question 6. (1 + tanθ + secθ) (1 + cotθ – cosecθ) is equal to (a) 0 (b) 1 (c) 2 (d) -1

Question 7. Ratios of sides of a right triangle with respect to its acute angles are known as (a) trigonometric identities (b) trigonometry (c) trigonometric ratios of the angles (d) none of these

Answer: (c) trigonometric ratios of the angles

Question 8. If tan θ = \(\frac { 12 }{ 5 }\), then \(\frac { 1+sinθ }{ 1-sinθ }\) is equal to (a) 24 (b) \(\frac { 12 }{ 13 }\) (c) 25 (d) 9

Answer: (c) 25

Question 9. The value of cos θ cos(90° – θ) – sin θ sin (90° – θ) is: (a) 1 (b) 0 (c) -1 (d) 2

Answer: (b) 0

Question 10. If x = a cos θ and y = b sin θ, then b 2 x 2 + a 2 y 2 = (a) ab (b) b 2 + a 2 (c) a 2 b 2 (d) a 4 b 4

Answer: (c) a 2 b 2

Question 11. If ΔABC is right angled at C, then the value of cos (A + B) is (a) 0 (b) 1 (c) \(\frac { 1 }{ 2 }\) (d) \(\frac { \sqrt { 3 } }{ 2 }\)

Answer: (a) 0

Question 12. If x and y are complementary angles, then (a) sin x = sin y (b) tan x = tan y (c) cos x = cos y (d) sec x = cosec y

Answer: (d) sec x = cosec y

Question 13. sin (45° + θ) – cos (45° – θ) is equal to (a) 2 cos θ (b) 0 (c) 2 sin θ (d) 1

Question 14. If 0° < θ < 90°, then sec 0 is (a) >1 (b) < 1 (c) =1 (d) 0

Answer: (a) >1

Question 15. In right triangle ABC, right angled at C, if tan A = 1, then the value of 2 sin A cos A is (a) 0 (b) 1 (c) – 1 (d) 2

Answer: (b) 1

Question 16. Given that sin A=\(\frac { 1 }{ 2 }\) and cos B=\(\frac { 1 }{ \sqrt { 2 } }\) then the value of (A + B) is: (a) 30° (b) 45° (c) 75° (d) 15°

Answer: (c) 75°

Question 17. If sin A = \(\frac { 1 }{ 2 }\), then the value of cot A is (a) √3 (b) \(\frac { 1 }{ \sqrt { 3 } }\) (c) \(\frac { \sqrt { 3 } }{ 2 }\) (d) 1

Answer: (a) √3

Question 18. If √3tanθ = 3sinθ, then the value of sin 2 θ−cos 2 θ is (a) 0 (b) 1 (c) \(\frac { 1 }{ 2 }\) (d) \(\frac { 1 }{ 3 }\)

Answer: (d) \(\frac { 1 }{ 3 }\)

Question 19. Out of the following options, the two angles that are together classified as complementary angles are (a) 120° and 60° (b) 50° and 30° (c) 65° and 25° (d) 70° and 30°

Answer: (c) 65° and 25°

Question 20. If sin θ − cos θ = 0, vthen the value of θ is (a) 90° (b) 30° (c) 45° (d) 60°

Answer: (c) 45°

Question 21. If tan 2A = cot (A – 18°), then the value of A is (a) 24° (b) 18° (c) 27° (d) 36°

Answer: (d) 36°

Question 22. If cos A + cos 2 A = 1, then sin 2 A + sin 4 A is (a) -1 (b) 0 (c) 1 (d) 2

Question 23. If sin θ + sin 2 θ = 1, then cos 2 θ + cos 4 θ = ____ (a) -1 (b) 0 (c) 1 (d) 2

Question 24. sin 2B = 2 sin B is true when B is equal to (a) 90° (b) 60° (c) 30° (d) 0°

Answer: (d) 0°

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  • Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

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Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry: Free PDF Download

CBSE Class 10 Maths Chapter 8 Important Questions revolve around the concept of trigonometric equations at its base. The Class 10 Maths Ch 8 Important Questions by Vedantu come with all the solutions that are drafted in a way, to provide you with a maximum understanding of the subject and figure out the different aspects of trigonometry. The important questions include questions from all the key concepts like basic trigonometry, opposite & adjacent sides in a right-angled triangle, basic trigonometric ratios, and standard values of trigonometric ratios and complementary trigonometric ratios. 

The solutions to these important questions are drafted in an easy-to-understand method. Additionally, the solutions contain step-wise explanations. Download the Class 10 important questions PDF to ensure a deeper understanding of the topic.

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Study Important Questions for Class 10 Mathematics Chapter 8 - Introduction to Trigonometry

1. If \[x\cos \theta y\sin \theta =a,\,x\sin \theta +y\cos \theta =b\], Prove that \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\].

\[x\cos \theta y\sin \theta =a\]        …… (1)

\[x\sin \theta +y\cos \theta =b\]        …… (2)

Squaring and adding the equation (1) and (2) on both sides.

\[{{x}^{2}}{{\cos }^{2}}\theta +{{y}^{2}}{{\sin }^{2}}\theta -2xy\cos \theta \sin \theta +{{x}^{2}}{{\sin }^{2}}\theta +{{y}^{2}}{{\cos }^{2}}\theta +2xy\cos \theta \sin \theta ={{a}^{2}}+{{b}^{2}}\]

\[\Rightarrow {{x}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  \right)+{{y}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta  \right)={{a}^{2}}+{{b}^{2}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]

\[\therefore {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]

Hence proved.

2. Prove that \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] Can Never Be Less Than \[2\].

Given: \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] 

We know that, ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\operatorname{cosec}}^{2}}\theta =1+{{\cot }^{2}}\theta $.

\[\Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =1+{{\tan }^{2}}\theta +1+{{\cot }^{2}}\theta \]

\[\Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =2+{{\tan }^{2}}\theta +{{\cot }^{2}}\theta \]

Therefore, \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] can never be less than \[2\].

3. If \[\sin \varphi =\dfrac{1}{2}\], show that \[3\cos \varphi -4{{\cos }^{3}}\varphi =0\].

Given: \[\sin \varphi =\dfrac{1}{2}\]

We know that $\sin 30{}^\circ =\dfrac{1}{2}$.

While comparing the angles of $\sin $, we get  

\[\Rightarrow \varphi ={{30}^{\circ }}\]

Substitute \[\varphi ={{30}^{\circ }}\] to get 

\[3\cos \varphi -4{{\cos }^{3}}\varphi =3\cos \left( 30{}^\circ  \right)-4{{\cos }^{3}}\left( 30{}^\circ  \right)\]

\[\Rightarrow 3\left( \dfrac{\sqrt{3}}{2} \right)-4\left( \dfrac{3\sqrt{3}}{8} \right)\Rightarrow 0\]

Therefore, \[3\cos \varphi -4{{\cos }^{3}}\varphi =0\] .

Hence proved .

4. If \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4\], then Show that \[\tan \varphi =\dfrac{1}{\sqrt{3}}\].

Given: \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4\]

We know that, ${{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi =1$ and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]

Then, \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4({{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi )\]

\[\Rightarrow 7{{\sin }^{2}}\varphi -4{{\sin }^{2}}\varphi =4{{\cos }^{2}}\varphi -3{{\cos }^{2}}\varphi \]

\[\Rightarrow 3{{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi \]

\[\Rightarrow \dfrac{{{\sin }^{2}}\varphi }{{{\cos }^{2}}\varphi }=\dfrac{1}{3}\]

\[\Rightarrow {{\tan }^{2}}\varphi =\dfrac{1}{3}\]

\[\Rightarrow \tan \varphi =\dfrac{1}{\sqrt{3}}\]

\[\therefore \tan \varphi =\dfrac{1}{\sqrt{3}}\]

5. If \[\cos \varphi +\sin \varphi =\sqrt{2}\cos \varphi \], Prove that \[\cos \varphi -\sin \varphi =\sqrt{2}\sin \varphi \].

Given : \[\cos \varphi +\sin \varphi =\sqrt{2}\cos \varphi \]

Squaring on both sides, we get

\[\Rightarrow {{\left( \cos \varphi +\sin \varphi  \right)}^{2}}=2{{\cos }^{2}}\varphi \]

\[\Rightarrow {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi +2\cos \varphi \sin \varphi =2{{\cos }^{2}}\varphi \]

$\Rightarrow {{\sin }^{2}}\varphi =2{{\cos }^{2}}\varphi -{{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi $

$\Rightarrow {{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi $

Add ${{\sin }^{2}}\varphi $ on both sides

\[\Rightarrow 2{{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi +{{\sin }^{2}}\varphi \]

\[\Rightarrow 2{{\sin }^{2}}\varphi ={{\left( \cos \varphi -\sin \varphi  \right)}^{2}}\] 

\[\therefore \cos \varphi -\sin \varphi =\sqrt{2}\sin \varphi \]

6. If \[\tan A+\sin A=m\] and \[\tan A-\sin A=n\], then Show that \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].

Given: 

\[\tan A+\sin A=m\]           …… (1)

\[\tan A-\sin A=n\]            …… (2)

Now to prove \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].

Take left-hand side

\[{{m}^{2}}-{{n}^{2}}={{\left( \tan A+\sin A \right)}^{2}}-{{\left( \tan A-\sin A \right)}^{2}}\]

\[\Rightarrow {{\tan }^{2}}A+{{\sin }^{2}}A+2\tan A\sin A-{{\tan }^{2}}A-{{\sin }^{2}}A+2\tan A\sin A\]

\[\Rightarrow 4\tan A\sin A\]         

$\therefore {{m}^{2}}-{{n}^{2}}=4\tan A\sin A$      …… (3)

Now take right-hand side

 \[4\sqrt{mn}=4\sqrt{\left( \tan A+\sin A \right)\left( \tan A-\sin A \right)}\]

\[\Rightarrow 4\sqrt{{{\tan }^{2}}A-{{\sin }^{2}}A}\Rightarrow 4\sqrt{\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-{{\sin }^{2}}A}\]

\[\Rightarrow 4\sqrt{{{\sin }^{2}}A\left( \dfrac{1}{{{\cos }^{2}}A}-1 \right)}\Rightarrow 4\sin A\sqrt{{{\sec }^{2}}A-1}\]

\[\Rightarrow 4\sin A\sqrt{{{\tan }^{2}}A}\Rightarrow 4\sin A\tan A\]

Hence, $4\sqrt{mn}=4\tan A\sin A$

\[\therefore {{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\]

Hence proved. 

7. If \[\sec A=x+\dfrac{1}{4x}\], then prove that \[\sec A+\tan A=2x\] or \[\dfrac{1}{2x}\].

Given: \[\sec A=x+\dfrac{1}{4x}\]

Squaring on both sides.

\[\Rightarrow {{\sec }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}\]

We know that, \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow 1+{{\tan }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}\]

\[\Rightarrow {{\tan }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}-1\]

\[\Rightarrow {{\tan }^{2}}A={{x}^{2}}+\dfrac{1}{16{{x}^{2}}}+\dfrac{1}{2}-1\Rightarrow {{x}^{2}}+\dfrac{1}{16{{x}^{2}}}-\dfrac{1}{2}\Rightarrow {{\left( x-\dfrac{1}{4x} \right)}^{2}}\]

Taking square root on both sides,

\[\Rightarrow \tan A=\pm \left( x-\dfrac{1}{4x} \right)\].

Now, find $\sec A+\tan A$

If $\tan A=x-\dfrac{1}{4x}$ means

$\sec A+\tan A=x+\dfrac{1}{4x}+x-\dfrac{1}{4x}\Rightarrow 2x$

$\therefore \sec A+\tan A=2x$

And if $\tan A=-x+\dfrac{1}{4x}$ means

$\sec A+\tan A=x+\dfrac{1}{4x}-x+\dfrac{1}{4x}\Rightarrow \dfrac{2}{4x}\Rightarrow \dfrac{1}{2x}$

$\therefore \sec A+\tan A=\dfrac{1}{2x}$

8. If \[A,B\]are Acute Angles and \[\sin A=\cos B\], then Find the Value of \[A+B\].

Given: $\sin A=\cos B$

We know that $\sin A=\cos \left( 90{}^\circ -A \right)$

While comparing the values to get

$\cos B=\cos \left( 90{}^\circ -A \right)$

$\Rightarrow B=90{}^\circ -A\Rightarrow A+B=90{}^\circ $

\[\therefore A+B={{90}^{\circ }}\].

9. Evaluate the Following Questions:

a. Solve for \[\phi \], if \[\tan 5\phi =1\].

Given: \[\tan 5\phi =1\]

We know that, ${{\tan }^{-1}}\left( 1 \right)=45{}^\circ $

$5\phi ={{\tan }^{-1}}\left( 1 \right)\Rightarrow 45{}^\circ $

$5\phi =45{}^\circ $

$\phi =\dfrac{45{}^\circ }{5}\Rightarrow 9{}^\circ $

\[\because \phi ={{9}^{\circ }}\]

b. Solve for \[\varphi \] , if \[\dfrac{\sin \varphi }{1+\cos \varphi }+\dfrac{1+\cos \varphi }{\sin \varphi }=4\].

Given: \[\dfrac{\sin \varphi }{1+\cos \varphi }+\dfrac{1+\cos \varphi }{\sin \varphi }=4\]

\[\dfrac{{{\sin }^{2}}\varphi +1+{{\cos }^{2}}\varphi +2\cos \varphi }{\sin \varphi \left( 1+\cos \varphi  \right)}=4\]

\[\dfrac{{{\sin }^{2}}\varphi +{{(1+\cos \varphi )}^{2}}}{\sin \varphi (1+\cos \varphi )}=4\]

\[\dfrac{2+2\cos \varphi }{\sin \varphi \left( 1+\cos \varphi  \right)}=4\]

\[\dfrac{2(1+\cos \varphi )}{\sin \varphi (1+\cos \varphi )}=4\]

\[\dfrac{2}{\sin \varphi }=4\]

\[\sin \varphi =\dfrac{1}{2}\]

We know that, $\sin 30{}^\circ =\dfrac{1}{2}$

\[\sin \varphi =\sin {{30}^{\circ }}\Rightarrow \varphi ={{30}^{\circ }}\]

\[\therefore \varphi ={{30}^{\circ }}\].

10. If \[\dfrac{\cos \alpha }{\cos \beta }=m\] and \[\dfrac{\cos \alpha }{\sin \beta }=n\], show that \[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\].

Given: \[\dfrac{\cos \alpha }{\cos \beta }=m\]         …… (1)

\[\dfrac{\cos \alpha }{\sin \beta }=n\]             …… (2)

Squaring equation (1) and (2). We get,

\[{{m}^{2}}=\dfrac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }\]

\[{{n}^{2}}=\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }\]

Now to prove \[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\] ,

Take left-hand side,  

\[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta =\left( \dfrac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }+\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[=\left( \dfrac{{{\cos }^{2}}\alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\alpha {{\cos }^{2}}\beta }{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[={{\cos }^{2}}\alpha \left( \dfrac{1}{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[=\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }\]

\[={{n}^{2}}\]

\[\therefore ({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\]

11. If \[7\cos ec\varphi -3\cot \varphi =7\], then prove that \[7\cot \varphi -3\cos ec\varphi =3\].

Given: \[7\cos ec\varphi -3\cot \varphi =7\]

Then prove that, \[7\cot \varphi -3\cos ec\varphi =3\]

\[7\cos ec\varphi -3\cot \varphi =7\]

\[49{{\operatorname{cosec}}^{2}}\varphi +9{{\cot }^{2}}\varphi -42\operatorname{cosec}\varphi \cot \varphi =49\]

We know that, ${{\operatorname{cosec}}^{2}}\varphi =1+{{\cot }^{2}}\varphi $ and ${{\cot }^{2}}\varphi ={{\operatorname{cosec}}^{2}}\varphi -1$.

\[49\left( {{\cot }^{2}}\varphi +1 \right)+9\left( {{\operatorname{cosec}}^{2}}\varphi -1 \right)-42\operatorname{cosec}\varphi \cot \varphi =49\]

\[49{{\cot }^{2}}\varphi +49+9{{\operatorname{cosec}}^{2}}\varphi -9-2\left( 3\operatorname{cosec}\varphi \cdot 7\cot \varphi  \right)=49\]

\[{{\left( 7\cot \varphi -3\operatorname{cosec}\varphi  \right)}^{2}}=49-49+9\]

\[{{\left( 7\cot \varphi -3\operatorname{cosec}\varphi  \right)}^{2}}=9\]

Take square root on both sides, we get

\[\therefore 7\cot \varphi -3\operatorname{cosec}\varphi =3\]

12. Prove that \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\].

Given: \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\]

Let us take left-hand side,

\[\begin{align}&2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1 \\ & =2\left( {{\left( {{\sin }^{2}}\varphi  \right)}^{3}}+{{\left( {{\cos }^{2}}\varphi  \right)}^{3}} \right)-3\left( {{\left( {{\sin }^{2}}\varphi  \right)}^{2}}+{{\left( {{\cos}^{2}}\varphi  \right)}^{2}} \right)+1 \\ \end{align}\]

\[=2\left[ {{\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)}^{3}}-3{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi \left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right) \right]-3\left[ {{\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)}^{2}}-2{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]+1\]

\[=2\left[ 1-3{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]-3\left[ 1-2{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]+1\]

\[=2-6{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi -3+6{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi +1\]

Therefore, \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\].

13. If \[\tan \theta =\dfrac{5}{6}\] and \[\theta =\phi ={{90}^{\circ }}\]. What is the value of \[\cot \phi \].

Given: \[\tan \theta =\dfrac{5}{6}\] and \[\theta =\phi ={{90}^{\circ }}\]

We know that, $\tan \theta =\dfrac{1}{\cot \theta }$.

$\cot \phi =\dfrac{1}{\tan \phi }$

$=\dfrac{1}{{5}/{6}\;}$

$=\dfrac{6}{5}$ 

$\therefore \cot \phi =\dfrac{6}{5}$.

14. What is the Value of \[\tan \varphi \] in terms of \[\sin \varphi \] ?

Given: \[\tan \varphi \]

We know that, \[\tan \varphi =\dfrac{\sin \varphi }{\cos \varphi }\] and \[{{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi =1\]

\[\tan \varphi =\dfrac{\sin \varphi }{\cos \varphi }\]

\[\therefore \tan \varphi =\dfrac{\sin \varphi }{\sqrt{1-{{\sin }^{2}}\varphi }}\]

15. If \[\sec \varphi +\tan \varphi =4\], Find the Value of \[\sin \varphi \], \[\cos \varphi \].

Given: \[\sec \varphi +\tan \varphi =4\]

\[\dfrac{1}{\cos \varphi }+\dfrac{\sin \varphi }{\cos \varphi }=4\]

\[\dfrac{1+\sin \varphi }{\cos \varphi }=4\]

$1+\sin \varphi =4\cos \varphi $

${{\left( 1+\sin \varphi  \right)}^{2}}={{\left( 4\cos \varphi  \right)}^{2}}$

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16{{\cos }^{2}}\varphi $

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16\left( 1-{{\sin }^{2}}\varphi  \right)$

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16-16{{\sin }^{2}}\varphi $

$17{{\sin }^{2}}\varphi +2\sin \varphi -15=0$

$17{{\sin }^{2}}\varphi +17\sin \varphi -15\sin \varphi -15=0$

$17\sin \varphi \left( \sin \varphi +1 \right)-15\left( \sin \varphi +1 \right)=0$

$\left( \sin \varphi +1 \right)\left( 17\sin \varphi -15 \right)=0$

If $\sin \varphi +1=0$

Hence, $\sin \varphi =-1$ is not possible.

Then, $17\sin \varphi -15=0$

$\therefore \sin \varphi =\dfrac{15}{17}$

Now find $\cos \varphi $

Substitute the value of $\sin \varphi $

\[1+\dfrac{15}{17}=4\cos \varphi \]

\[\dfrac{32}{17}=4\cos \varphi \]

\[\Rightarrow \cos \varphi =\dfrac{32}{17\left( 4 \right)}\Rightarrow \dfrac{8}{17}\]

\[\therefore \cos \varphi =\dfrac{8}{17}\]

Short Answer Questions (2 Marks)

1. In \[\Delta ABC\] , Right Angled at \[B,AB=24cm,BC=7cm\].

Determine the Following Equations:    

(i) \[\sin A,\cos A\]

Let us draw a right-angled triangle \[ABC\], right angled at \[B\].

A right triangle ABC with AB=24cm and BC=7cm

Using Pythagoras theorem, find $AC$.

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[={{(24)}^{2}}+{{(7)}^{2}}\]

\[=576+49\]

\[\therefore AC=25cm\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{7}{25}\]

\[\therefore \sin A=\dfrac{7}{25}\]

 \[\cos A=\dfrac{AB}{AC}=\dfrac{24}{25}\]

\[\therefore \cos A=\dfrac{24}{25}\]

(ii) \[\sin C,\cos C\]

A right triangle ABC

\[\sin C=\dfrac{AB}{AC}=\dfrac{24}{25}\]       

\[\therefore \sin C=\dfrac{24}{25}\]

\[\cos C=\dfrac{BC}{AC}=\dfrac{7}{25}\]

\[\therefore \cos C=\dfrac{7}{25}\]

2. In Adjoining Figure, Find the Value of \[\tan P-\cot R\].     

A right triangle ABC with AB=24cm and BC=7cm - (2)

Using Pythagoras theorem,

$P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}$

\[{{(13)}^{2}}={{(12)}^{2}}+Q{{R}^{2}}\]

\[Q{{R}^{2}}=169-144\Rightarrow 25\]

\[\therefore QR=5cm\]

Then find \[\tan P-\cot R\] ,

First find the value of $\tan P$

$\tan P=\dfrac{side\,opp.\,to\,\angle P}{side\,adj.\,to\,\angle P}=\dfrac{QR}{PQ}=\dfrac{5}{12}$

$\therefore \tan P=\dfrac{5}{12}$

Now find the value of $\cot R$

We know that, $\tan R=\dfrac{1}{\cot R}$

For that we need to first find the value of $\tan R$

$\tan R=\dfrac{side\,opp.\,to\,\angle R}{side\,adj.\,to\,\angle R}=\dfrac{PQ}{QR}=\dfrac{12}{5}$

$\therefore \cot R=\dfrac{5}{12}$

Then, 

\[\tan P-\cot R=\dfrac{QR}{PQ}-\dfrac{QR}{PQ}\]

\[\Rightarrow \dfrac{5}{12}-\dfrac{5}{12}\Rightarrow 0\]

\[\therefore \tan P-\cot R=0\].

3. If \[\sin A=\dfrac{3}{4}\], Calculate the Value of \[\cos A\] and \[\tan A\].                                                         

A right triangle ABC with AC=4k and BC=3k

Given that the triangle \[ABC\] in which \[\angle B={{90}^{\circ }}\]

Let us take \[BC=3k\] and \[AC=4k\]

Then using Pythagoras theorem,

\[AB=\sqrt{{{(AC)}^{2}}-{{(BC)}^{2}}}\]

 \[\Rightarrow \sqrt{{{(4k)}^{2}}-{{(3k)}^{2}}}\]

 \[\Rightarrow \sqrt{16k-9k}\]

\[\Rightarrow k\sqrt{7}\]

$\therefore AB=k\sqrt{7}$

Calculate the value of \[\cos A\]  

$\cos A=\dfrac{AB}{AC}=\dfrac{k\sqrt{7}}{4k}=\dfrac{\sqrt{7}}{4}$

$\therefore \cos A=\dfrac{\sqrt{7}}{4}$

And calculate the value of \[\tan A\]

$\tan A=\dfrac{BC}{AB}=\dfrac{3k}{k\sqrt{7}}=\dfrac{3}{\sqrt{7}}$

$\therefore \tan A=\dfrac{3}{\sqrt{7}}$

4. Given \[15\cot A=8\], Find the Values of \[\sin A\] and \[\sec A\].                                                           

Given: \[15\cot A=8\]

Let us assume a triangle \[ABC\] in which \[\angle B={{90}^{\circ }}\]. 

\[15\cot A=8\]

\[\Rightarrow \cot A=\dfrac{8}{15}\]

Since $\cot A=\dfrac{adj}{hyp}=\dfrac{AB}{BC}$.

Let us draw the triangle.

A right triangle ABC with AB=8k and BC=15k

Now, \[AB=8k\] and \[BC=15k\].

Using Pythagoras theorem, find the value of $AC$.

\[AC=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}}\]

\[\Rightarrow \sqrt{{{(8k)}^{2}}+{{(15k)}^{2}}}\]

$\Rightarrow \sqrt{64{{k}^{2}}+225{{k}^{2}}}$

$\Rightarrow \sqrt{289{{k}^{2}}}$

\[\Rightarrow 17k\]

$\therefore AC=17k$

Now, find the values of \[\sin A\] and \[\sec A\] .

\[\sin A=\dfrac{BC}{AC}=\dfrac{15k}{17k}=\dfrac{15}{17}\]

$\therefore \sin A=\dfrac{15}{17}$

\[\sec A=\dfrac{AC}{AB}=\dfrac{17k}{8k}=\dfrac{17}{8}\]

 \[\therefore \sec A=\dfrac{17}{8}\]

5. If \[\angle A\] and \[\angle \,B\] are Acute Angles Such That \[\cos A=\cos B\], then show that \[\angle A=\angle \,B\]

A right triangle ABC right angled at C

Given: \[\cos A=\cos B\]

In right triangle \[ABC\],

\[\cos A=\dfrac{side\,adj.\,A}{hyp.}=\dfrac{AC}{AB}\]  …… (1)

And, \[\cos B=\dfrac{side\,adj.\,B}{hyp.}=\dfrac{BC}{AB}\] …… (2)

Then, \[\cos A=\cos B\]

Now, equate equation (1) and (2).

\[\Rightarrow \dfrac{AC}{AB}=\dfrac{BC}{AB}\]

$\Rightarrow AC=BC$

\[\Rightarrow \angle A=\angle B\].

Therefore, Angles opposite to equal sides are equal.

6. State Whether the Following are True or False. Justify Your Answer.           

 The Value of \[\tan A\] is Always Less than \[1\].

False because sides of a right triangle may have any length, so \[\tan A\] may have any value. For example, $\tan A=\dfrac{BC}{AB}=\dfrac{15}{10}=\dfrac{3}{2}=1.5$. 

  \[\sec A=\dfrac{12}{5}\] for Some Value of Angle \[A\] .

True as \[\sec A\] is always greater than \[1\]. For example, $\sec A=\dfrac{hyp.}{side\,adj.\,A}$ . As hypotenuse will be the largest side. So, it is true.

  \[\cos A\] is the Abbreviation Used for the Cosecant of Angle \[A\] .

False as \[\cos A\] is the abbreviation of \[\operatorname{cosineA}\]. Because $\cos A$ means cosine of angle $A$ and $\operatorname{co}\sec A$ means cosecant of angle $A$.

 \[\cot A\] is the Product of \[\cot \]and \[A\].

False as \[\cot A\] is not the product of \[cot\] and \[A\]. \[cot\] without \[A\] doesn’t have meaning. 

 \[\sin \theta =\dfrac{4}{3}\] for Some Angle \[\theta \].

Ans:  

False as \[\sin \theta \] cannot be greater than \[1\]. For example, $\sin \theta =\dfrac{side\,opp.\,\theta }{hyp}$. Since the hypotenuse is the largest side. So, \[\sin \theta \] will be less than $1$.

7. Evaluate the Following Equations: 

i. \[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}\]

Given: \[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}\]

We know that, $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $

\[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}=\dfrac{\sin ({{90}^{\circ }}-{{72}^{\circ }})}{\cos {{72}^{\circ }}}=\dfrac{\cos {{72}^{\circ }}}{\cos {{72}^{\circ }}}=1\]                   

$\therefore \dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=1$

ii. \[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}\]

Given: \[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}\]

We know that, $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $

\[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}=\dfrac{\tan ({{90}^{\circ }}-{{64}^{\circ }})}{\cot {{64}^{\circ }}}=\dfrac{\cot {{64}^{\circ }}}{\cot {{64}^{\circ }}}=1\]

$\therefore \dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=1$

iii. \[\cos {{48}^{\circ }}-\sin {{42}^{\circ }}\]

Given: \[\cos {{48}^{\circ }}-\sin {{42}^{\circ }}\]

We know that, $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $

\[\Rightarrow \cos ({{90}^{\circ }}-{{42}^{\circ }})-\sin {{42}^{\circ }}\]

\[\Rightarrow \sin {{42}^{\circ }}-\sin {{42}^{\circ }}\Rightarrow 0\]

 \[\therefore \cos 48{}^\circ -\sin 42{}^\circ =0\]

iv. \[\operatorname{cosec}31{}^\circ -\sec 59{}^\circ \]

Given: \[\operatorname{cosec}31{}^\circ -\sec 59{}^\circ \]

We know that, $\operatorname{cosec}\left( 90{}^\circ -\theta  \right)=\sec \theta $

\[\Rightarrow \operatorname{cosec}(({{90}^{\circ }}-{{59}^{\circ }})-\sec {{59}^{\circ }}\]

\[\Rightarrow \sec {{59}^{\circ }}-\sec {{59}^{\circ }}\Rightarrow 0\]

\[\therefore \operatorname{cosec}31{}^\circ -\sec 59{}^\circ =0\]

1. Show that the Following Equations:                       

\[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

Given: \[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

We know that, $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $.

Now let us take left-hand side,

\[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \tan ({{90}^{\circ }}-{{42}^{\circ }})\tan ({{90}^{\circ }}-{{67}^{\circ }})\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \cot {{42}^{\circ }}\cot {{67}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \dfrac{1}{\tan {{42}^{\circ }}}.\dfrac{1}{\tan {{67}^{\circ }}}.\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow 1\] is equal to R.H.S

\[\therefore \tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

(ii) \[\cos {{38}^{\circ }}\cos {{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0\]

Given: \[\cos {{38}^{\circ }}\cos {{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0\]

Now let us take left-hand side, 

\[\cos {{38}^{\circ }}cos{{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow \cos ({{90}^{\circ }}-{{52}^{\circ }})\cos ({{90}^{\circ }}-{{38}^{\circ }})-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow \sin {{52}^{\circ }}\sin {{38}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow 0\] is equal to R.H.S

$\therefore \cos {{38}^{\circ }}cos{{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0$

2. If  \[\tan 2A=\cot (A-{{18}^{\circ }})\] where \[2A\] is an Acute Angle, Find the Value of \[A\]. 

Given: \[\tan 2A=\cot (A-{{18}^{\circ }})\]

We know that, $\cot \left( 90{}^\circ -\theta  \right)=\tan \theta $

\[\Rightarrow \cot ({{90}^{\circ }}-2A)=\cot (A-{{18}^{\circ }})\]

Now equalise the angles,

\[{{90}^{\circ }}-2A=A-{{18}^{\circ }}\]

\[-2A-A=-{{18}^{\circ }}-{{90}^{\circ }}\]

\[-3A=-{{108}^{\circ }}\]

\[A=\dfrac{108{}^\circ }{3}\]

$\therefore A=36{}^\circ $

3. If \[\tan A=\cot B\], then Prove That \[A+B={{90}^{\circ }}\].                                                            

Given: \[\tan A=\cot B\]

\[\cot ({{90}^{\circ }}-A)=\cot B\]

\[{{90}^{\circ }}-A=B\]

\[\Rightarrow A+B={{90}^{\circ }}\]

$\therefore A+B=90{}^\circ $

4. If \[\sec 4A=\operatorname{cosec}(A-{{20}^{\circ }})\], Where \[4A\] is an Acute Angle, Then Find the Value of $A$.

Ans:  

Given: \[\sec 4A=\operatorname{cosec}(A-{{20}^{\circ }})\]

\[\Rightarrow \operatorname{cosec}({{90}^{\circ }}-4A)=\operatorname{cosec}(A-{{20}^{\circ }})\]

\[{{90}^{\circ }}-4A=A-{{20}^{\circ }}\]

\[-4A-A=-{{20}^{\circ }}-{{90}^{\circ }}\]

\[-5A=-{{110}^{\circ }}\]

\[A=\dfrac{{{110}^{\circ }}}{5}\]

\[\therefore A={{22}^{\circ }}\]

5. If \[A,B\] and \[C\] are Interior Angles of a \[\Delta ABC\], then Show That \[\sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}\].

Given: \[A,B\] and \[C\] are interior angles of a \[\Delta ABC\].

We know that, \[A+B+C={{180}^{\circ }}\].

Let us consider, 

\[\dfrac{A+B+C}{2}={{90}^{\circ }}\]

\[\Rightarrow \dfrac{B+C}{2}={{90}^{\circ }}-\dfrac{A}{2}\]

Multiply $\sin $ on both sides,

\[\sin \left( \dfrac{B+C}{2} \right)=\sin \left( {{90}^{\circ }}-\dfrac{A}{2} \right)\]

\[\therefore \sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}\]

6. Express \[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}\] in terms of trigonometric ratios of angles between  \[{{0}^{\circ }}\] and \[{{45}^{\circ }}\].                                                   

Given : \[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}\].

We know that, $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $ and $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $.

\[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}=\sin ({{90}^{\circ }}-{{23}^{\circ }})+\cos ({{90}^{\circ }}-{{15}^{\circ }})\]

\[=\cos {{23}^{\circ }}+\sin {{15}^{\circ }}\]

\[\therefore \cos {{23}^{\circ }}+\sin {{15}^{\circ }}\] is the required value.

7. Express the Trigonometric Ratios \[\sin A,\sec A\] and \[\tan A\] in Terms of \[\cot A\]. 

Find the value of \[\sin A\] in terms of \[\cot A\].

By using identity \[{{\operatorname{cosec}}^{2}}A-{{\cot }^{2}}A=1\].

Then, use $\operatorname{cosec}A=\dfrac{1}{\sin A}$

\[\Rightarrow {{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A\]

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}A}=1+{{\cot }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

\[\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

Now find the value for \[\sec A\] in terms of \[\cot A\].

Using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}\]

\[\Rightarrow {{\sec }^{2}}A=\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}\]

\[\Rightarrow \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\] 

\[\therefore \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

And find the value for \[\tan A\] in terms of \[\cot A\]

By trigonometric ratio property, $\tan A=\dfrac{1}{\cot A}$

Hence, \[\tan A=\dfrac{1}{\cot A}\]

Therefore, \[\sin A,\sec A\] and \[\tan A\] are founded in terms of \[\cot A\].  

8. Write the Other Trigonometric Ratios of $A$ in Terms of \[\sec A\] .                        

Find the value of \[\sin A\] in terms of \[\sec A\]

By using identity,\[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=1-\dfrac{1}{{{\sec }^{2}}A}\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{{{\sec }^{2}}A-1}{{{\sec }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

\[\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

Now find the value for \[\cos A\] in terms of \[\sec A\],

By trigonometric ratio property, $\cos A=\dfrac{1}{\sec A}$

$\therefore \cos A=\dfrac{1}{\sec A}$

Find the value for $\tan A$ in terms of \[\sec A\],

By using identity, \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\tan }^{2}}A={{\sec }^{2}}A-1\]

\[\Rightarrow \tan A=\sqrt{{{\sec }^{2}}A-1}\]

\[\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}\]

Find the value for \[\operatorname{cosec}A\] in terms of \[\sec A\]

By trigonometric ratio property, \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]

\[\Rightarrow \operatorname{cosec}A=\dfrac{1}{\sin A}\]

Substitute the value of \[\sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\].

\[\Rightarrow \operatorname{cosec}A=\dfrac{1}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}\]

\[\Rightarrow \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

\[\therefore \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

Finally, find the value for \[\cot A\] in terms of \[\sec A\]

By trigonometric ratio property, \[\cot A=\dfrac{1}{\tan A}\]

$\Rightarrow \cot A=\dfrac{1}{\tan A}$

Substitute the value of \[\tan A=\sqrt{{{\sec }^{2}}A-1}\]

\[\Rightarrow \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$.

9. Evaluate the Following Equations:                                    

(i) \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

Given: \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and  \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]                      

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$         

(ii) \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

Given: \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

We know that, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$

\[\Rightarrow \sin \left( 25{}^\circ +65{}^\circ  \right)\]

\[\Rightarrow \sin 90{}^\circ \]                    

\[\Rightarrow 1\]

$\therefore \sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}=1$

10. Show that Any Positive Odd Integer Is of the Form \[\mathbf{6q}\text{ }+\text{ }\mathbf{1}\], or \[\mathbf{6q}\text{ }+\text{ }\mathbf{3},\]or \[\mathbf{6q}\text{ }+\text{ }\mathbf{5}\], where \[q\] is some integer.                                                      

Let \[a\] be any positive integer and \[b=\text{ }6\]. 

Then, by Euclid’s algorithm,

\[a=6q+r\] for some integer \[q\ge 0\], and \[r=0,1,2,3,4,5\] because \[0\le r<\text{ }6\].

Therefore, \[a=6q\] or \[6q+\text{ }1\] or \[6q+\text{ }2\]or \[6q\text{ }+3\]or \[6q+\text{ }4\]or \[6q+\text{ }5\]

Also, \[6q+1=2\times 3q+1=2{{k}_{1}}+1,\] where \[{{k}_{1}}\] is a positive integer

\[6q+3=(6q+2)+1=2(3q+1)+1=2{{k}_{2}}+\text{ }1,\]Where \[{{k}_{2}}\]is an integer

\[6q+5=(6q+4)+1=2(3q+2)+1=2{{k}_{3}}+1\], where \[{{k}_{3}}\] is an integer

Clearly, \[6q+1,6q+3,6q+5\] are of the form \[2k+\text{ }1,\]where \[k\]an integer is.

Therefore, \[6q+1,6q+3,6q+5\] are not exactly divisible by \[2\].

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form \[6q+1\], or \[6q+3\], or\[6q+5.\]

11. An Army Contingent of \[\mathbf{616}\] Members are to March Behind an Army Band of \[\mathbf{32}\] Members in a Parade. The Two Groups Are to March in the Same Number of Columns. What Is the Maximum Number of Columns in Which They Can March?

We have to find the \[HCF\left( 616,\text{ }32 \right)\] to find the maximum number of columns in which they can march. 

To find the HCF, we can use Euclid’s algorithm.

\[616=32\times 19+8\]

\[\Rightarrow 32=8\times 4+0\]

Hence,  \[HCF\left( 616,\text{ }32 \right)\] is \[8\].

Therefore, they can march in \[8\] columns each.

12. Use Euclid’s Division Lemma to Show That the Square of Any Positive Integer Is Either of Form \[\mathbf{3}m\]or \[\mathbf{3}m+\mathbf{1}\] for some integer \[m\]. 

[Hint: Let \[x\] be any positive integer then it is of the form\[\mathbf{3}q,\mathbf{3}q+\mathbf{1}\] or \[\mathbf{3}q+\mathbf{2}\text{ }\]. Now square each of these and show that they can be rewritten in the form \[\mathbf{3}m\]or \[\mathbf{3}m+\mathbf{1}\].]

Let\[\text{ }a\] be any positive integer and \[b=3\].

Then \[a=3q+r\] for some integer \[q\ge 0\]

And \[r=0,1,2\] because \[0\le r<3\]

Therefore, \[a=3q\] or \[3q+1\] or \[3q+2\]

\[\Rightarrow {{a}^{2}}={{(3q)}^{2}}\] or \[{{(3q+1)}^{2}}\] or \[{{(3q+2)}^{2}}\]

\[\Rightarrow {{a}^{2}}={{(9q)}^{2}}\] or \[9{{q}^{2}}+6q+1\] or \[9{{q}^{2}}+12q+4\]

\[\Rightarrow {{a}^{2}}=3\times {{(3q)}^{2}}\] or \[3(3{{q}^{2}}+2q)+1\] or \[3(3{{q}^{2}}+4q+1)+1\]

\[\Rightarrow a=3{{k}_{1}}\] or \[3{{k}_{2}}+1\] or \[3{{k}_{3}}+1\]

Where \[{{k}_{1}},{{k}_{2}},{{k}_{3}}\] are some positive integers 

Hence, it can be said that the square of any positive integer is either of the form \[3m\]or \[3m+1\].

Short Answer Questions (3 Marks)

1. Given \[\sec \theta =\dfrac{13}{12}\] , Calculate the Values for All Other Trigonometric Ratios. 

A triangle ABC right angled at B and angle A is 𝛳

Given:   \[\sec \theta =\dfrac{13}{12}\]

Let us consider a triangle \[ABC\] in which \[\angle A=\theta \] and \[\angle B={{90}^{\circ }}\]

Let \[AB=12k\] and \[AC=13k\]

Then, find the value of $BC$

\[BC=\sqrt{{{(AC)}^{2}}-{{(AB)}^{2}}}\]

\[\begin{align}&\Rightarrow\sqrt{{{(13k)}^{2}}-{{(12k)}^{2}}}\\&\Rightarrow \sqrt{69{{k}^{2}}-144{{k}^{2}}} \\ & \Rightarrow \sqrt{25{{k}^{2}}} \\ & \Rightarrow 5k \\ \end{align}\]

\[\therefore BC=5k\]

Since, \[\sec \theta =\dfrac{13}{12}\]

Similarly, 

\[\sin \theta =\dfrac{BC}{AC}=\dfrac{5k}{13k}=\dfrac{5}{13}\]

\[\cos \theta =\dfrac{AB}{AC}=\dfrac{12k}{13k}=\dfrac{12}{13}\]

\[\tan \theta =\dfrac{BC}{AB}=\dfrac{5k}{12k}=\dfrac{5}{12}\]

\[\cot \theta =\dfrac{AB}{BC}=\dfrac{12k}{5k}=\dfrac{12}{5}\]

\[\cos ec\theta =\dfrac{AC}{BC}=\dfrac{13k}{5k}=\dfrac{13}{5}\]

2. If \[\cot \theta =\dfrac{7}{8}\] , then Evaluate the Followings Equations :                                                                                        

i. \[\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

Given: \[\cot \theta =\dfrac{7}{8}\]

Then, \[AB=7k\] and \[BC=8k\]

Using Pythagoras theorem, find $AC$

\[AC=\sqrt{{{(BC)}^{2}}+{{(AB)}^{2}}}\]

\[\Rightarrow \sqrt{{{(8k)}^{2}}+{{(7k)}^{2}}}\]

\[\Rightarrow \sqrt{64{{k}^{2}}+49{{k}^{2}}}\]

\[\Rightarrow \sqrt{113{{k}^{2}}}\]

\[\Rightarrow \sqrt{113}k\]

\[\therefore AC=\sqrt{113}k\]

A triangle ABC right angled at B and angle A is 𝛳, sides AB=7k, BC=8k and AC=113k

Now find the value of trigonometric ratios.

\[\sin \theta =\dfrac{BC}{AC}=\dfrac{8k}{\sqrt{113}k}=\dfrac{8}{\sqrt{113}}\] and \[\cos \theta =\dfrac{AB}{AC}=\dfrac{7k}{\sqrt{113}k}=\dfrac{7}{\sqrt{113}}\].

\[\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

We know that, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$

\[\Rightarrow \dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\Rightarrow \dfrac{49}{113}\cdot \dfrac{113}{64}\Rightarrow \dfrac{49}{64}\]

$\therefore \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{49}{64}$

ii. \[{{\cot }^{2}}\theta \]

Given: \[{{\cot }^{2}}\theta \]

We know that, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

\[\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\Rightarrow \dfrac{49}{64}\]

$\therefore {{\cot }^{2}}\theta =\dfrac{49}{64}$

Hence, $\dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}$ and ${{\cot }^{2}}\theta $ are same.

3. If \[3\cot A=4\], then show that  \[\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\].              

A triangle ABC right angled at B and sides AB=4k, BC=3k and AC=5k

Given: \[3\cot A=4\]

Let us consider a triangle \[ABC\] in which  \[\angle B={{90}^{\circ }}\]

Then, \[3\cot A=4\] 

\[\Rightarrow \cot A=\dfrac{4}{3}\]

Let \[AB=4k\] and \[BC=3k\]

 \[\Rightarrow \sqrt{{{(3k)}^{2}}+{{(4k)}^{2}}}=\sqrt{16{{k}^{2}}+9{{k}^{2}}}\]

\[\Rightarrow \sqrt{25{{k}^{2}}}=5k\]

\[\therefore AC=5k\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{3k}{5k}=\dfrac{3}{5}\]  , \[\cos A=\dfrac{AB}{AC}=\dfrac{4k}{5k}=\dfrac{4}{5}\] and \[\tan A=\dfrac{BC}{AB}=\dfrac{3k}{4k}=\dfrac{3}{4}\].

To prove: \[\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\]

Let us take left-hand side 

L.H.S \[=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}\]

Substitute the value of $\tan A$.

\[\Rightarrow \dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}=\dfrac{16}{25}-\dfrac{9}{25}\]

\[\Rightarrow \dfrac{7}{25}\]

\[\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{7}{25}\]

R.H.S \[={{\cos }^{2}}A-{{\sin }^{2}}A\]

\[\Rightarrow {{\left( \dfrac{4}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{16}{25}-\dfrac{9}{25}\]

$\therefore {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{7}{25}$.

It shows that \[\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}\]

\[\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\]

4. In \[\Delta ABC\] Right Angles at \[B\], if \[A=\dfrac{1}{\sqrt{3}}\], then Find Value of the Following Equations:                                  

(i) \[\sin A\cos C+\cos A\sin C\]

A triangle ABC right angled at B and sides AB=3k, BC=k and AC=2k

Let \[BC=k\]and \[AB=\sqrt{3}k\]

Then, using Pythagoras theorem find $AC$

\[\Rightarrow \sqrt{{{(k)}^{2}}+{{(\sqrt{3}k)}^{2}}}\Rightarrow \sqrt{{{k}^{2}}+3{{k}^{2}}}\Rightarrow \sqrt{4{{k}^{2}}}\Rightarrow 2k\]

\[\therefore AC=2k\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{k}{2k}=\dfrac{1}{2}\] and \[\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}k}{2k}=\dfrac{\sqrt{3}}{2}\]

For \[\angle C\], adjacent \[=BC\], opposite \[=AB\], and hypotenuse \[=AC\]

\[\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}k}{2k}=\dfrac{\sqrt{3}}{2}\] and \[\cos A=\dfrac{BC}{AC}=\dfrac{k}{2k}=\dfrac{1}{2}\]

Now find the values of the following equations,

\[\sin A\cos C+\cos A\sin C\]

\[\Rightarrow \dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{1}{4}+\dfrac{3}{4}\Rightarrow \dfrac{4}{4}\Rightarrow 1\]

\[\therefore \sin A\cos C+\cos A\sin C=1\]

(ii) \[\cos A\cos C-\sin A\sin C\]

\[\Rightarrow \dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}\Rightarrow 0\]

\[\therefore \cos A\cos C-\sin A\sin C=0\]

5. In \[\Delta PQR\], right angled at \[Q\],\[PR+QR=25cm\] and \[PQ=5cm\]. Determine the Values of  \[\sin P,\cos P\]  and \[\tan P\].                                                                      

A triangle PQR right angled at Q and sides PQ=5cm, QR= x cm and PR=(25-x) cm

Given: In \[\Delta PQR\], right angled at \[Q\]

And \[PR+QR=25cm\], \[PQ=5cm\]

Let us take \[QR=xcm\] and \[PR=(25-x)cm\]

By using Pythagoras theorem, find the value of $x$.

\[R{{P}^{2}}=R{{Q}^{2}}+Q{{P}^{2}}\]

\[\Rightarrow {{(25-x)}^{2}}={{(x)}^{2}}+{{(5)}^{2}}\Rightarrow 625-50x+{{x}^{2}}={{x}^{2}}+25\]

\[\Rightarrow -50x=-600\Rightarrow x=12\]

Hence, \[RQ=12cm\]and \[RP=25-12=13cm\]

Now, find the values of \[\sin P,\cos P\]   and \[\tan P\].     

\[\therefore \sin P=\dfrac{RQ}{RP}=\dfrac{12}{13}\], \[\cos P=\dfrac{PQ}{RP}=\dfrac{5}{13}\] and \[\tan P=\dfrac{RQ}{PQ}=\dfrac{12}{5}\].

6. If \[\tan (A+B)=\sqrt{3}\] and \[\tan (A-B)=\dfrac{1}{\sqrt{3}}\]; \[{{0}^{\circ }}<A+B\le {{90}^{\circ }}\]; \[A>B\]. Find \[A\]and \[B\].  

Given : $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$.

We know that,$\tan 60{}^\circ =\sqrt{3}$ and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

$\tan \left( A+B \right)=\tan 60{}^\circ $

$\Rightarrow A+B=60{}^\circ $          …… (1)

$\tan \left( A-B \right)=\tan 30{}^\circ $

$\Rightarrow A-B=30{}^\circ $          …… (2)

Adding equation (1) and (2). We get,

$\begin{align} & A+B+A-B=60{}^\circ +30{}^\circ \\ & \Rightarrow 2A=90{}^\circ \Rightarrow A=45{}^\circ \\ \end{align}$

$\therefore A=45{}^\circ $

Put $A=45{}^\circ $ in equation (1).

$A+B=60{}^\circ $

$\Rightarrow 45{}^\circ +B=60{}^\circ \Rightarrow B=60{}^\circ -45{}^\circ \Rightarrow B=15{}^\circ $

$\therefore B=15{}^\circ $

Hence, $A=45{}^\circ $ and $B=15{}^\circ $.

7. Choose the Correct Option. Justify Your Choice:   

(i) \[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]\[=\]

\[1\] 

\[9\] 

\[8\] 

Ans: (B) $9$

\[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]

\[\Rightarrow 9({{\sec }^{2}}A-{{\tan }^{2}}A)\Rightarrow 9\times 1\Rightarrow 9\]

(ii) \[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]\[=\]

\[0\] 

\[2\] 

none of these

Ans: (C) \[2\]

\[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]

\[\Rightarrow \left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)\Rightarrow \left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{(\cos \theta +\sin \theta )}^{2}}-{{(1)}^{2}}}{\cos \theta .\sin \theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\]        

\[\Rightarrow \dfrac{1+2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\Rightarrow \dfrac{2\cos \theta \sin \theta }{\cos \theta .\sin \theta }\Rightarrow 2\]

(iii) \[(\sec A+\tan A)(1-\sin A)\]\[=\]

\[\sec A\] 

\[\sin A\] 

\[\cos ecA\] 

Ans : (D) \[\cos A\]

\[(\sec A+\tan A)(1-\sin A)\]

\[\Rightarrow \left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)(1-\sin A)\Rightarrow \left( \dfrac{1+\sin A}{\cos A} \right)(1-\sin A)\]

We know that, \[1-{{\sin }^{2}}A={{\cos }^{2}}A\]

\[\Rightarrow \dfrac{1-{{\sin }^{2}}A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A}{\cos A}\Rightarrow \cos A\]  

(iv) \[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\]

\[{{\sec }^{2}}A\] 

\[-1\] 

\[{{\cot }^{2}}A\] 

Ans: (D) ${{\tan }^{2}}A$

\[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sec }^{2}}A-{{\tan }^{2}}A+{{\tan }^{2}}A}{\cos e{{c}^{2}}A-{{\cot }^{2}}A+{{\cot }^{2}}A}\]

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfac{rac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

\[\Rightarrow \dfr{{\sin }^{2}}A}{{{\cos }^{2}}A}\Rightarrow {{\tan }^{2}}A\].

Long Answer Questions (4 Marks)

1. Express the Trigonometric Ratios \[\sin A,\sec A\] and \[\tan A\] in Terms of \[\cot A\].  

Find the value for \[\sin A\] in terms of \[\cot A\]

By using identity \[\cos e{{c}^{2}}A-{{\cot }^{2}}A=1\]

$\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$

Express the value of \[\sec A\] in terms of \[\cot A\]

By using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\sec }^{2}}A=\dfrac{1+{{\cot }^{2}}A}{{{\cot }^{2}}A}\]

\[\Rightarrow \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

Express the value of \[\tan A\] in terms of \[\cot A\]

We know that, $\tan A=\dfrac{1}{\cot A}$

\[\therefore \tan A=\dfrac{1}{\cot A}\]

2. Write the Other Trigonometric Ratios of \[A\] in Terms of \[\sec A\] .                 

Express the value of \[\sin A\] in terms of \[\sec A\]

By using identity, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\Rightarrow 1-\dfrac{1}{{{\sec }^{2}}A}\Rightarrow \dfrac{{{\sec }^{2}}A-1}{{{\sec }^{2}}A}\]

Express the value of \[\cos A\] in terms of \[\sec A\]

We know that, \[\cos A=\dfrac{1}{\sec A}\]

\[\therefore \cos A=\dfrac{1}{\sec A}\]

Express the value of \[\tan A\] in terms of \[\sec A\]

$\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}$

Express the value of \[\cos ecA\] in terms of \[\sec A\]

We know that, $\operatorname{cosec}A=\dfrac{1}{\sin A}$

Substitute the value of \[\sin A\]

\[\Rightarrow \cos ecA=\dfrac{1}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}\Rightarrow \dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}$

Express the value of \[\cot A\] in terms of \[\sec A\]

We know that, \[\cot A=\dfrac{1}{\tan A}\]

Substitute the value of \[\tan A\]

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$

3. Evaluate the following equations:                                                                                                       

(i). \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]                   

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$

(ii). \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

\[\Rightarrow \sin {{25}^{\circ }}\cos ({{90}^{\circ }}-{{25}^{\circ }})+\cos {{25}^{\circ }}\sin ({{90}^{\circ }}-{{25}^{\circ }})\]

\[\Rightarrow \sin {{25}^{\circ }}.\sin {{25}^{\circ }}+\cos {{25}^{\circ }}.\cos {{25}^{\circ }}\]                   

\[\Rightarrow {{\sin }^{2}}{{25}^{\circ }}+{{\cos }^{2}}{{25}^{\circ }}=1\]      

\[\therefore \sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}=1\]

4. Choose the Correct Option. Justify Your Choice:  

\[\Rightarrow \dfrac{1-{{\sin }^{2}}A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A}{\cos A}\Rightarrow \cos A\] 

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\Rightarrow {{\tan }^{2}}A\].

5. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                      

(i). ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Given: ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

We know that, \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

Then, let us take left-hand side

\[\Rightarrow {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}={{\operatorname{cosec}}^{2}}\theta +{{\cot }^{2}}\theta -2\operatorname{cosec}\theta \cot \theta \] 

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}\theta }+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-2\times \dfrac{1}{\sin \theta }.\dfrac{\cos \theta }{\sin \theta }\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-\dfrac{2\cos \theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{1+{{\cos }^{2}}\theta -2\cos \theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{{{(1-\cos \theta )}^{2}}}{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

\[\Rightarrow \dfrac{1-\cos \theta }{1+\cos \theta }=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

(ii) \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

Given: \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

\[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A+1+{{\sin }^{2}}A+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A+1+2\sin A}{(1+\sin A)\cos A}\]      

\[\Rightarrow \dfrac{1+1+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2(1+\sin A)}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2}{\cos A}\Rightarrow 2\sec A=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

(iii). \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Given: \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

We know that, \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take L.H.S

\[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }\times \dfrac{\sin \theta }{\sin \theta -\cos \theta }+\dfrac{\cos \theta }{\sin \theta }\times \dfrac{\cos \theta }{\cos \theta -\sin \theta }\]

\[\Rightarrow \dfrac{{{\sin }^{2}}\theta }{\cos \theta (\sin \theta -\cos \theta )}-\dfrac{{{\cos }^{2}}\theta }{\sin \theta (\sin \theta -\cos \theta )}\]

\[\Rightarrow \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta (\sin \theta -\cos \theta )}\]                     

\[\Rightarrow \dfrac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{1}{\sin \theta \cos \theta }+1\]

\[\Rightarrow 1+\dfrac{1}{\sin \theta \cos \theta }\Rightarrow 1+\sec \theta \cos ec\theta \] = R.H.S

$\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $

(iv) \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Given: \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Then, let us take L.H.S 

\[\dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{\cos A+1}{\cos A}\times \dfrac{\cos A}{1}\Rightarrow 1+\cos A\]

\[\Rightarrow 1+\cos A\times \dfrac{1-\cos A}{1-\cos A}\Rightarrow \dfrac{1-{{\cos }^{2}}A}{1-\cos A}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{1-\cos A}=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$

(v) \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\], using the identity \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

Given: \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\]

We know that, \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

\[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}\]

Dividing all terms by \[\sin A\]

\[\Rightarrow \dfrac{\cot A-1+\cos ecA}{\cot A+1-\cos ecA}\]

\[\Rightarrow \dfrac{\cot A+\cos ecA-1}{\cot A-\cos ecA+1}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)-(\cos e{{c}^{2}}A-{{\cot }^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)+({{\cot }^{2}}A-\cos e{{c}^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)(1+\cot A-\cos ecA)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \cot A+\cos ecA=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A$

(vi) \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

Given: \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] and \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

\[\sqrt{\dfrac{1+\sin A}{1-\sin A}}\]

Let us take conjugate of the term. Then,

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}\times \sqrt{\dfrac{1+\sin A}{1+\sin A}}\]

\[\Rightarrow \sqrt{\dfrac{{{(1+\sin A)}^{2}}}{1-{{\sin }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{{{\operatorname{Cos}}^{2}}A}}\]

\[\Rightarrow \dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]                           

\[\Rightarrow \sec A+\tan A=\text{R}\text{.H}\text{.S}\]           

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$       

Hence proved.        

(vii) \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

Given: \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]

\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2{{\cos }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta \left[ 2(1-{{\sin }^{2}}\theta )-1 \right]}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2-2{{\sin }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}{\cos \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\tan \theta =\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta $

(viii) \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

Given: \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

We know that, \[\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta \] and \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]

\[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}={{\left( \sin A+\dfrac{1}{\sin A} \right)}^{2}}+{{\left( \cos A+\dfrac{1}{\cos A} \right)}^{2}}\]

\[={{\sin }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+2\sin A.\dfrac{1}{\sin A}+{{\cos }^{2}}A+\dfrac{1}{{{\cos }^{2}}A}+2\cos A.\dfrac{1}{\cos A}\]

\[=2+2+{{\sin }^{2}}A+{{\cos }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=4+1+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=5+\cos e{{c}^{2}}A+{{\sec }^{2}}A\]

\[=5+1+{{\cot }^{2}}A+1+{{\tan }^{2}}A\]

\[=7+{{\tan }^{2}}A+{{\cot }^{2}}A=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$

(ix) \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

Given: \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

\[(\cos ecA-\sin A)(\sec A-\cos A)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)\]

\[=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)\]

\[=\dfrac{{{\cos }^{2}}A}{\sin A}\times \dfrac{{{\sin }^{2}}A}{\cos A}\]

\[=\sin A.\cos A\]

\[=\dfrac{\sin A.\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}\]

Dividing all the terms by \[\sin A.\cos A\]

\[=\dfrac{\dfrac{\sin A.\cos A}{\sin A.\cos A}}{\dfrac{{{\sin }^{2}}A}{\sin A.\cos A}+\dfrac{{{\cos }^{2}}A}{\sin A.\cos A}}\]

\[=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}\]

\[=\dfrac{1}{\tan A+\cot A}=\text{R}\text{.H}\text{.S}\]

$\therefore (\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}$

(x) \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

Given:\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

We know that, \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] and \[1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}A\]

\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)=\dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\]

\[=\dfrac{1}{{{\cos }^{2}}A}\times \dfrac{{{\sin }^{2}}A}{1}\]

\[={{\tan }^{2}}A=\text{R}\text{.H}\text{.S}\]

Now, prove the Middle side 

\[{{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}}\]

\[={{\left( \dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}} \right)}^{2}}\]

\[=\left( \dfrac{1-\tan A}{\dfrac{-(1-\tan A)}{\tan A}} \right)\]

\[={{(-\tan A)}^{2}}\]

$\therefore \left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$

6. Use Euclid’s Division Algorithm to Find the \[\text{HCF}\] of:                                     

(i) \[\mathbf{135}\] and \[\mathbf{225}\]

Given: \[135\] and \[225\]

We have \[225>135\],

So, we have to apply the division lemma to \[225\] and \[135\] to obtain

\[225=135\times 1+90\]

Here remainder\[90\ne 0\], again we are applying the division lemma to \[135\] and \[90\] to obtain 

\[135=90\times 1+45\]

Again, the remainder\[45\ne 0\], then apply the division lemma to obtain 

\[90=2\times 45+0\]

Since now we got remainder as zero. Here, the process get stops. 

The divisor at this stage is \[45\]

Therefore, the HCF of \[135\] and \[225\] is \[45\].

(ii) \[\mathbf{196}\] and\[~\mathbf{38220}\] 

Given: \[196\] and \[38220\]

We have \[38220>196\], 

So, we have to apply the division lemma to \[38220\] and \[196\] to obtain 

\[38220=196\times 195+0\]

Since we get the remainder as zero, the process stops here. 

The divisor at this stage is \[196\], 

Therefore, HCF of \[196\] and \[38220\] is \[196\].  

(iii) \[~\mathbf{867}\] and \[\mathbf{255}\]

Given: \[867\] and \[255\]

We have 867 > 255, 

So, we have to apply the division lemma to \[867\]and \[255\] to obtain 

\[867=255\times 3+102\]

Here remainder \[102\ne 0\], again apply the division lemma to\[255\] and \[102\] to obtain 

\[255=102\times 2\text{ }51\]

Again, remainder \[51\ne 0\], again apply the division lemma to \[102\] and \[51\] to obtain 

\[102=51\times 2+0\]

The divisor at this stage is \[51\], 

Therefore, HCF of \[867\] and \[255\] is \[51\].

7. Evaluate the Following Equations:                                                                                                              

i. \[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

Given: \[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

We know that, $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2},\,\sin 30{}^\circ =\dfrac{1}{2}$ and $\cos 60{}^\circ =\dfrac{1}{2},\,\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

\[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

\[=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}\]

\[=\dfrac{3}{4}+\dfrac{1}{4}\]

\[=\dfrac{4}{4}=1\]

\[\therefore \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}=1\]

ii. \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

Given: \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

We know that, $\tan 45{}^\circ =1$, $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

Then, \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

\[=2{{(1)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}\]

\[=2+\dfrac{3}{4}-\dfrac{3}{4}=2\]

$\therefore 2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}=2$

iii. \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}\]

Given: \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}\]

We know that, $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\,\sec 30{}^\circ =\dfrac{2}{\sqrt{3}}$ and $\operatorname{cosec}30{}^\circ =2$

\[=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}\]

\[=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}\]

\[=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\]

\[=\dfrac{\sqrt{3}}{\sqrt{2}\times 2(\sqrt{3}+1)}\]

\[=\dfrac{\sqrt{3}}{\sqrt{2}\times 2(\sqrt{3}+1)}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}\]

\[=\dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{2}\times 2(3-1)}\]

\[=\dfrac{\sqrt{3}(\sqrt{3}-1)}{4\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}\]\[\]

\[=\dfrac{3\sqrt{2}-\sqrt{6}}{8}\]

$\therefore \dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}=\dfrac{3\sqrt{2}-\sqrt{6}}{8}$

(iv) \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

Given: \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

We know that,  $\sin 30{}^\circ =\dfrac{1}{2},\,\tan 45{}^\circ =1,\,\operatorname{cosec}60{}^\circ =\dfrac{2}{\sqrt{3}},\,\sec 30{}^\circ =\dfrac{2}{\sqrt{3}},\,\cos 60{}^\circ =\dfrac{1}{2}$ and $\cot 45{}^\circ =1$.

Then, \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

$=\dfrac{{1}/{2}\;+1-{2}/{\sqrt{3}}\;}{{2}/{\sqrt{3}}\;+{1}/{2}\;+1}$

\[=\dfrac{\dfrac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}\]

\[=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\]

\[=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \dfrac{3\sqrt{3}-4}{3\sqrt{3}-4}\]

\[=\dfrac{27+16-24\sqrt{3}}{27-16}\]

\[=\dfrac{43-24\sqrt{3}}{11}\]

$\therefore \dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}=\dfrac{43-24\sqrt{3}}{11}$

(v) \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}\]

Given: \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}\]. Then,

\[=\dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{3}{\sqrt{3}} \right)}^{2}}-{{(1)}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}\]

\[=\dfrac{5\times \dfrac{1}{4}+4\times \dfrac{4}{3}-1}{\dfrac{1}{4}+\dfrac{3}{4}}\]

\[=\dfrac{\dfrac{1}{12}\times 67}{\dfrac{4}{4}}\]

\[=\dfrac{67}{12}\]

$\therefore \dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}=\dfrac{67}{12}$

8. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                                      

(i) ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta  }{1+\cos \theta }$

(iii) \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Given: \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

Key Topics Covered in Chapter 8

Chapter 8 of Class 10 Mathematics consists of the discussion of the following key concepts.

Basic Trigonometry

Opposite & Adjacent Sides In A Right-Angled Triangle

Basic Trigonometric Ratios

Standard Values Of Trigonometric Ratios 

Complementary Trigonometric Ratios 

Significance of Important Questions of Trigonometry Class 10

Trigonometry is one of the most significant parts of Important Questions for Class 10 Maths Chapter 8. This chapter revolves around the memorization and conceptual understanding of the user’s problem-solving ability of a given space, preferably a triangle. These Class 10 Maths Trigonometry Important Questions help students to have a better understanding and application of trigonometry in the real world and clear the base around its introductory stage, to help students retain its benefits in the long run.

Did You Know?

Trigonometry has different applications and a few of them are mentioned below:

It is used in cartography for the creation of maps.

It has applications in the aviation industry and satellite systems.

It is also used to describe light and sound waves. 

This was the complete discussion on CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry important questions. If you are a Class 10 student we highly recommend downloading and practising the important questions on trigonometry. Practising these important questions will not only help you understand the concepts but will also help you in analysing the important topics and exam patterns. 

We wish you all the very best! Be exam ready with Vedantu!

Important Related Links for CBSE Class 10 Maths

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FAQs on Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

1. Is Studying Chapter 8 Maths Class 10 Important Questions Helpful in Preparing for the Exams?

Studying Trigonometry Important Questions Class 10 helps you uncover the different aspects of trigonometry and teach the subject at its core. The pool of experienced teachers who have a complete idea of the marking schematics and all the necessary questions that can come in the exams are focused on providing the students with a complete kit that helps them secure maximum marks in their upcoming exams. Further, the step by step explanation of practical problems is what set our solutions apart, making it completely reliable a solution to start preparing for your exams.

2. What are the important questions in Chapter 8 Trigonometry of Class 10 Maths?

Class 10 Maths Chapter 8 is related to trigonometry and its functions. Important questions given on Vedantu for trigonometry Class 10 Chapter 8 can help students to prepare for their exams. Important questions are given according to the latest syllabus and pattern of CBSE. All questions are prepared using sample papers, previous year papers, and the latest trends in the CBSE board exams. Students can use important questions in trigonometry Class 10 to prepare for their final board exams. These solutions are available on Vedantu's official website(vedantu.com) and mobile app free of cost.

3. What is the easiest way to solve Chapter 8 Trigonometry of Class 10 Maths questions?

Students have to understand the basic concepts of Trigonometry for solving Class 10 Maths Chapter 8 questions. They have to memorize the different identities properly for solving the NCERT questions quickly. Students can also refer to the important questions in Class 10 Maths Chapter 8 available for free on Vedantu to understand the basic concepts and formulas used for solving trigonometry easily.  They can visit Vedantu to download important questions for Class 10 Maths Chapter 8 for regular practice.

4. How many exercises are there in Chapter 8 Trigonometry of Class 10 Maths?

In Trigonometry Class 10 Maths Chapter 8, there are four exercises. All exercises are based on different concepts related to trigonometry. Students need to solve all questions given in four exercises in the NCERT book to understand the concepts of trigonometry and for scoring high marks in Class 10 maths board exams. Students can practice solving trigonometry from the NCERT Solutions available on Vedantu in the offline and online mode. They can download the NCERT solutions for Class 10 Maths Chapter 8 on computers to work offline.

5. What are the different topics studied in Chapter 8 Trigonometry of Class 10 Maths?

The different topics studied in Class 10 Maths Chapter 10 include an introduction to trigonometry,  information about trigonometric ratios, trigonometric ratios of specific angles, trigonometric ratios of complementary angles, and trigonometric identities. Students will learn all topics step by step in each exercise of the chapter. They can practice NCERT Solutions given on different topics of trigonometry on Vedantu. NCERT Solutions for trigonometry Class 10 are prepared by experts to help students understand the concepts and score high marks. 

CBSE Class 10 Maths Important Questions

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Class 10 Maths Chapter 18 Trigonometry Important Questions

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In Class 10 ICSE (Indian Certificate of Secondary Education) mathematics, Chapter 2 may not specifically cover the topic of "Banking" as its primary focus. Instead, the curriculum typically includes topics related to commercial mathematics, which can encompass banking concepts. Commercial mathematics includes areas such as simple interest, compound interest, and profit and loss, which are essential aspects of banking and finance. Here are trigonometric identities class 10 ICSE important questions.

Introduction

What is trigonometry, class 10 trigonometry important questions and answers, icse class 10 maths chapter wise important questions, frequently asked questions.

In Class 10 ICSE (Indian Certificate of Secondary Education) mathematics, the chapter on "Trigonometry" is a fundamental topic that explores the relationships between angles and sides of right triangles. Trigonometry is a branch of mathematics that has wide-ranging applications, including in navigation, engineering, physics, and more. Here's an introduction to trigonometry in Class 10 ICSE mathematics: Importance of Trigonometry: Trigonometry is a critical branch of mathematics because it helps us solve real-world problems involving angles and distances. It's especially valuable in fields that require precise measurements and calculations. Common Trigonometric Formulas: Trigonometric Identities: These include the Pythagorean identities (sin^2θ + cos^2θ = 1) and various angle sum and difference identities. Special Angles: Trigonometry deals with special angles like 30 degrees, 45 degrees, and 60 degrees, which have well-defined trigonometric values.

Ans: Trigonometry is the study of the relationships between the angles and sides of triangles. It primarily focuses on right triangles, which have one angle equal to 90 degrees. Trigonometric functions and ratios, such as sine, cosine, and tangent, are used to establish these relationships. Key Concepts and Objectives: Trigonometric Ratios: The fundamental trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). They relate the angles of a right triangle to the lengths of its sides. Right Triangles: Trigonometry primarily deals with right triangles, where one angle is 90 degrees. The side opposite the right angle is the hypotenuse, and the other two sides are the legs. Sine, Cosine, and Tangent: These trigonometric ratios are defined as follows: Sine (sin): Opposite side / Hypotenuse Cosine (cos): Adjacent side / Hypotenuse Tangent (tan): Opposite side / Adjacent side

trigonometry class 10 icse important questions

Q1. \(\frac{cos A}{1-sin A}\) - tan A =

(a) cos a (b) sec a (c) sin a (d) cosec a.

Explanation: \(\frac{cos A}{1- sin A}\) - tan A = \(\frac{cos A(1+sin A)}{1-sin A(1+ sin A)}\)- tan A =\(\frac{cos A(1+ sin A)}{(1+sin^2 A)}\)-tan A =\(\frac{cos A(1+ sin A)}{cos^2 A}\)- tan A =\(\frac{(1+sin A)}{cos A}\)-tan A =\(\frac{1}{cos A}\)+\(\frac{sin A}{cos A}\)- tan A

Q2. \(\frac{cos A}{1-sin A}\)-tan A=

(a) cos a (b) sec a (c) sin a (d) cosec a.

Ans . (b) Explanation: \(\frac{cos A}{1-sin A}\)-tan A = \(\frac{cos A(1+ sin A)}{(1- sin A)(1+ sin A)}\)- tan A =\(\frac{cos A(1+sin A)}{1+sin^2 A}\)- tan A =\(\frac{cos A(1+ sin A)}{cos ^2 A}\)- tan A =\(\frac{1+ sin A}{cos A}\)- tan A =\(\frac{1}{cos A}\)+\(\frac{sin A}{cos A}\)- tan A = sec A + tan A - tan A  = sec A.

Q3. Prove that : (i)\(\sqrt{\frac{1-cos\theta}{1+cos \theta}}\)= cosec θ - cos θ (ii) \(\sqrt{\frac{1+cos \theta}{1-cos \theta}}\)= sec θ - tan θ

Explanation: (i) L.H.S. =\(\sqrt{\frac{1-cos\theta}{1+cos\theta}×\frac{1-cos\theta}{1-cos\theta}}\) =\(\sqrt{\frac{(1-cos\theta)^2}{1+cos^2\theta}}\) =\(\frac{1-cos\theta}{\sqrt{1-cos^2\theta}}\) =\(\frac{1-cos\theta}{\sqrt{sin^2\theta}}\) =\(\frac{1-cos \theta}{sin \theta}\) =\(\frac{1}{sin \theta}-\frac{cos \theta}{sin \theta}\) = cosec θ – cot θ = R.H.S. Hence Proved. (ii)L.H.S.= \(\sqrt{\frac{1+sin \theta}{1-sin\theta}×\frac{1+sin \theta}{1+sin\theta}}\) =\(\sqrt{\frac{(1+sin\theta)^2}{1-sin^2\theta}}\) =\(\frac{1+sin\theta}{cos\theta}\)=\(\frac{1}{cos\theta}\)+\(\frac{sin\theta}{cos \theta}\) = sec θ + tan θ = R.H.S. Hence Proved

Q4. Prove that : sin 4 θ – cos 4 θ = sin 2 θ – cos 2 θ = 2 sin 2 θ – 1 = 1 – 2 cos 2 θ.

Explanation: Consider, sin 4 θ – cos 4 θ = (sin 2 θ) 2 – (cos 2 θ) 2 = (sin 2 θ – cos 2 θ)(sin 2 θ + cos 2 θ) [∵ (a – b)(a + b) = a 2 – b 2 ] = (sin 2 θ – cos 2 θ) × 1 [∵ sin 2 θ+ cos 2 θ= 1] = sin 2 θ – cos 2 θ = sin 2 θ – (1 – sin 2 θ) [∵cos 2 θ = 1 – sin 2 θ] = sin 2 θ – 1 + sin 2 θ = 2 sin 2 θ – 1 = 2(1 – cos 2 θ) – 1 [∵ sin 2 θ =1 – cos 2 θ] = 2 – 2 cos 2 θ – 1 = 1 – 2 cos 2 θ. Hence Proved.

Q5. The angle of elevation of a cloud from a point 200 metres above a lake is 30° and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud.

Explanation: Let P be the point of observation and C, the position of cloud. CN perpendicular from C on the surface of the lake and C‘ be the reflection of the cloud in the lake so that

Trigonometry_Q5

CN = NC´ = x (say) Then, PM = 200 m ∴   AN = MP = 200 m CA = CN – AN = (x – 200) m C´A = NC´ + AN = (x + 200) m Let PA = y m Then, in right angled Δ  PAC, \(\frac{CA}{PA}\)= tan 30 ⟹ \(\frac{x+200}{y}\)=\(\sqrt{3}\) ⇒ x + 200 = \(\sqrt{3}\)y ⟹y = \(\frac{y+200}{\sqrt{3}}\) ...( ii ) From equations (i) and (ii), \(\frac{x+200}{y}\) =\(\sqrt{3}\)(x-200) ⇒        x +200 = 3( x –200) ⇒        x +200 = 3 x –600 ⇒        2 x = 800 ⇒        x = 400 m ‍ Hence, the height of the cloud = 400 m.

trigonometry class 10 icse important questions

The study of trigonometry in Class 10 ICSE mathematics is essential for understanding the relationships between angles and sides in right triangles. It equips students with tools to solve practical problems and lays the foundation for more advanced mathematics and scientific applications. If you seek additional practice and a deeper comprehension of the topics covered in the chapter, oswal.io offers an extensive array of class 10 Trigonometry important questions and answers to facilitate a more profound understanding of the concepts.

Q1 : What is trigonometry, and why is it important?

Ans: Trigonometry is the study of the relationships between angles and sides in triangles. It's important because it helps us solve real-world problems involving measurements, distances, and angles.

Q2: What are the primary trigonometric ratios?

Ans:  The primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan).

Q3 : What is the sine of an angle in a right triangle?

Ans: The sine (sin) of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

Q4 : What is the cosine of an angle in a right triangle?

Ans: The cosine (cos) of an angle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

Q5 : How is the tangent of an angle calculated in a right triangle?

Ans: The tangent (tan) of an angle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.

Chapter Wise  Important Questions for ICSE Board Class 10 Mathematics

trigonometry class 10 icse important questions

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  • CBSE Class 10
  • Class 10 Maths Objective Questions
  • Chapter 9 Applications Of Trigonometry Objective Questions

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Objective Questions

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Objective Questions deal with how trigonometry helps to find the height and distance of different objects without any measurement. In earlier days, astronomers used trigonometry to calculate the distance from the earth to the planets and stars. Trigonometry is also mostly used in navigation and geography to locate the position in relation to the latitude and longitude. Keeping in mind the latest modification of the CBSE exam pattern, we have compiled the CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Objective Questions for the students to prepare well for the board exams. Students will learn the applications of trigonometry with real-life examples in a better way with the help of the MCQs provided in this article.

Solving these CBSE Class 10 Maths Chapter 9 objective questions is a good practice as it provides additional questions for the students to practise for the board exams. Students also get acquainted with a wide range of questions of different types and difficulty levels by practising these objective questions.

Find below the CBSE Class 10 Maths Objective Questions and the sub-topics it covers.

List of Sub-topics Covered in Chapter 9

We have compiled here some 20 multiple-choice questions covering the below-given topics from the chapter.

9.1 Applications of Trigonometry (6 MCQs from the Topic)

9.2 Introduction (4 MCQs from the Topic)

9.3 Heights and Distances (10 MCQs from the Topic)

Download Free CBSE Class 10 Maths Chapter 9 – Applications of Trigonometry Objective Questions PDF

Applications of trigonometry.

  • A technician has to repair a light on a pole of a height of 10 m. She needs to reach a point 1 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use when inclined at an angle of 60 ∘ to the ground that would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder?

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 1 Options

Answer: (D) 6√3 m

Solution: The given situation is explained in the figure below:

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 1 Solution- 1

DB – DC = CB

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 1 Solution-2

∴ The height of the ladder should be 6√3 m.

  • A statue, 2 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60°, and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 2 Options

Answer: (B) 2(√3 – 1)

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 2 Solution- 1

AB = x (using tan 45∘)

BD = AB √3 = x √3

We know that CD = BD – BC = 2m

X = 2(√3 -1 )

Therefore, the height of the pedestal is 2(√3 -1 ).

Answer: (C) 20 m

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 3 Solution-1

The given situation is explained in the figure above.

tan60°=DC/BC⇒BC=DC/tan60°= 60/ √3 = 20 √3 m

tan30°=AB/BC

⇒AB=BC×tan30°=20 √3 × (1/√3 m) = 20 m

Thus, the height of the building is 20 m.

Answer: (B) 20 m

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 4 Solution -1

The above figure represents the situation given in the question.

tan60°=AB/ BC

⇒AB=BCtan60°= BC √3……….(1)

⇒BC=AB/tan60°=AB/ tan60°

tan30°=AB/BD=AB/ (CD+BC)

⇒DC+BC=AB/tan30°=AB √3

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 4 Solution-2

Hence, the distance between the opposite bank of the canal is 20 m.

  • As observed from the top of a 150 m high lighthouse from the sea level, the angles of depression of the two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

cbse class 10 maths chapter 9 question 5 options

Answer: ( B) 150 (√3 – 1)

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 5 Solution - 1

Here, the lighthouse, BC = 150 m

BD = BC = 150 m (using tan 45°)

AB = BC √3 (using tan 30°)

So, distance AD = 150 (√3 – 1)

Hence, the between the two ships is 150 (√3 – 1).

Answer: (B) 30°

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 6 Solution- 1

Height of the pole that is above man’s height = 2 √3 – √3 = √3 m

Hence, AB = √3 m

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 6 Solution-4

⟹ C, angle of elevation = 30°

Therefore, the angle of elevation of the top of the pole from the observer is 30°.

Introduction

  • The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Applications of Trigonometry Objective Questions-1

Solution: In ΔABC, taking the tangent of ∠C, we have,

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 7 Solution - 1

tan C=AB/BC

⇒tan 30°=h/30

⇒ \(\begin{array}{l}\frac{1}{\sqrt{3}}\end{array} \) = h/30

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 7 Solution- 3

=17.32m = 10√3 m

Hence, the height of the tower is 10√3 metres.

Answer: (D) 30 m

Solution: Let AB be the tower of height h, and CD be the observer of height, 1.5 m at a distance of 28.5 m from tower AB.

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 8 Solution -1

In ΔAED, we have

tan45°=h/28.5

∴ h=AE+BE=AE+DC

= (28.5+1.5) m=30 m

Height of tower = h + 1.5

= 28.5 + 1.5

Hence, the height of the tower is 30 m.

  • An electrician has to repair an electrical fault on a pole of a height of 4 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use when inclined at an angle of 60° to the horizontal that would enable him to reach the required position?

Applications of Trigonometry Objective Questions-2

Solution: Let AC be the electric pole of height 4 m. Let B be a point 1.3 m below the top A of the pole AC.

Then, BC = AC – AB = (4 – 1.3) m = 2.7 m

Let BD be the ladder inclined at an angle of 60° to the horizontal.

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 9 Solution Image 1

In ΔBCD, we have

sin60°=2.7/L.

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 9 Solution Image 2

Answer: (A) 10 m

Solution: Let AB be the vertical pole and CA be the 20 m long rope, such that one end is tied from the top of the vertical pole AB and the other end C is tied to a point C on the ground.

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 10 Solution Image 1

In ΔABC, we have

sin 30°=h/20

Hence, the height of the pole is 10 m.

Heights and Distances

Answer: (C) 45 m

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 11 Solution Image 1

The given situation is explained in the figure above:

In triangle ABE,

tan45°=AB/EB

Also, EB=DC

∴ tan45°=AB/ DC

⇒AB=DC × tan 45°

⇒AB=1×42.75

Hence, the height of the chimney = AC = AB + BC

We can observe that BC = ED.

Thus, AC = AB + ED

= 42.75 + 2.25

Hence, the height of the chimney is 45 m.

  • A tower stands vertically on the ground. From a point on the ground, which is 30 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 30°. Find the height of the tower.

Applications of Trigonometry Objective Questions-3

Answer: (B) 10√3 m

Solution: The given situation is explained in the Δ below:

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 12 Solution - 1

Now, tan30°=AB/ BC

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 12 Solution - 2

∴ The height of the tower is  10√3 m.

  • The angles of depression of the top and the bottom of a 10 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building.

Applications of Trigonometry Objective Questions-4

The above figure represents the situation aptly.

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 13 Solution Image 2

(ED is the height of the building)

Answer: (A) 4

Solution: The given situation can be explained using the figure below:

CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Question 14 Solution Image 1

In the right-angled triangle ABC,

sin∠ABC=AC/AB=1/2

⇒sin30°=2/AB⇒AB= 2/ (1/2)

∴ The length of the slide is 4m.

  • A kite is flying at a height of 30 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Applications of Trigonometry Objective Questions-4

Solution: The situation can be explained using the figure below:

CBSE Class 10 Maths Chapter 9 Question 15 Solution Image 1

In the given right-angled triangle:

sin (∠ACB) =AB/AC

⇒sin60°=AB/AC

CBSE Class 10 Maths Chapter 9 Question 15 Solution Image 2

∴ The length of the string is 20√3 m

  • A vertical pole of 30 m is fixed on a tower. From a point on the level ground, the angles of elevation of the top and bottom of the pole are 60° and 45°. Find the height of the tower.

Applications of Trigonometry Objective Questions-5

Answer: (B) 15 (√3 + 1)

CBSE Class 10 Maths Chapter 9 Question 16 Solution Image 1

The situation can be explained by the figure above.

CBSE Class 10 Maths Chapter 9 Question 16 Solution Image 2

Therefore, the height of the tower is 15 (√3 + 1).

  • The value of tan A +sin A=M, and tan A – sin A=N.

The value of (M 2 −N 2 ) / (MN) 0.5

Solution: M 2 -N 2  = (Tan A+ Sin A + Tan A –Sin A) (Tan A +Sin A – Tan A+ Sin A)

M 2 -N 2  =4 tan A sin A

And (MN) 0.5  = (tan 2 A−sin 2 A) 0.5

CBSE Class 10 Maths Chapter 9 Question 17 Solution Image 1

  • Two towers, A and B, are standing at some distance apart. From the top of tower A, the angle of depression of the foot of tower B is found to be 30°. From the top of tower B, the angle of depression of the foot of tower A is found to be 60°. If the height of tower B is ‘h’ m, then the height of tower A in terms of ‘h’ is _____ m

Applications of Trigonometry Objective Questions-6

Answer: (B) h/3 m

CBSE Class 10 Maths Chapter 9 Question 18 Solution Image 1

Let the height of tower A be = AB = H.

And the height of tower B = CD = h

In triangle ABC,

tan30° = AB/AC = H/AC ……………………………. 1

In triangle ADC,

tan60° = CD/AC = h/AC………………………………….2

Divide 1 by 2

We get tan30°/tan60° = H/h

Therefore, H = h/3 m.

Answer: (D) 34.55 m

CBSE Class 10 Maths Chapter 9 Question 19 Solution Image 1

The above figure explains the situation given in the question.

Applications of Trigonometry Objective Questions-7

∴ The distance moved by the boy, d, is 34.55 m.

Answer: (D) 73.2 m

Solution: Let C and D be the objects and CD be the distance between the objects.

CBSE Class 10 Maths Chapter 9 Question 20 Solution Image 1

In ΔABC, tan 45° = AB/AC

AB=AC=100 m

In ΔABD, tan 30° = AB/AD

CBSE Class 10 Maths Chapter 9 Question 20 Solution Image 2

CD=AD−AC=173.2−100=73.2 metres

Hence, the distance between the two objects is 73.2 metres.

You can access the PDF link to download multiple-choice questions from Chapter 9 – Applications of Trigonometry, categorised as per the topics from which it is taken. Solving these questions will help the students to score better in the board exams because, as per the latest exam pattern, more objective questions are expected to be included in question papers.

CBSE Class 10 Maths Chapter 9 Extra MCQs

1. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, calculate the angle of elevation of the cloud. 

(a) equal to the angle of depression of its reflection.

(b) double to the angle of depression of its reflection

(c) not equal to the angle of depression of its reflection

(d) information insufficient

2. There are two windows in the house. A window of the house is at a height of 1.5 m above the ground, and the other window is 3 m vertically above the lower window. Ram and Shyam are sitting inside the two windows. At an instant, the angle of elevation of a balloon from these windows is observed as 45° and 30°, respectively. Then, what is the height of the balloon from the ground?

(a)  7.269 m

(b) 8.269 m

(c)  7.598 m

(d) 8.598 m

Apart from these MCQs, students can also download other resources at BYJU’S like NCERT Solutions , previous years’ papers, Syllabus , study notes, sample papers , and important questions for all the classes to prepare more efficiently for the exams. Keep learning and stay tuned for further updates on CBSE .

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Class 10 Math MCQ PDF (Questions and Answers) | 10th Grade Math MCQs Book Download

Quiz questions chapter 1-13 & practice tests with answers key | class 10 math textbook notes, mcqs & study guide, publisher description.

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Practice Algebraic Formulas and Applications MCQs with Answers PDF, chapter 2 test to solve MCQ questions: Algebraic expressions, Math formulas, surds and surds applications. Practice Algebraic Manipulation MCQs with Answers PDF, chapter 3 test to solve MCQ questions: Basic operations on algebraic fractions, square root of algebraic expression, HCF, and LCM. Practice Basic Statistics MCQs with Answers PDF, chapter 4 test to solve MCQ questions: Construction of frequency polygon, construction of histograms, frequency distribution, measures of central tendency, and measures of dispersion. Practice Factorization MCQs with Answers PDF, chapter 5 test to solve MCQ questions: Factorization of expressions, and Math theorems. Practice General Math MCQs with Answers PDF, chapter 6 test to solve MCQ questions: Basic concepts, circle's basic concepts, fraction, improper fraction, proper fraction, rational fraction, Math theorems, parallel lines, relation between roots, and coefficients. 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Practice Matrices and Determinants MCQs with Answers PDF, chapter 10 test to solve MCQ questions: Introduction to matrices, types of matrices, addition and subtraction of matrices, multiplication of matrices, multiplicative inverse of matrix, and solution of simultaneous linear equations. Practice Ratio, Proportions and Variations MCQs with Answers PDF, chapter 11 test to solve MCQ questions: Ratios, proportion, variation, joint variation, k method, and Math theorems. Practice Roots of Quadratic Equations MCQs with Answers PDF, chapter 12 test to solve MCQ questions: Quadratic equation, solution of quadratic equations complex cube roots of unity, discriminant, radical equation, and reciprocal equation. Practice Sets and Functions MCQs with Answers PDF, chapter 13 test to solve MCQ questions: Sets, recognize of operations on sets, example of sets, binary relation, and ordered pairs.

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CBSE Class 10 Hindi exam analysis: ‘Questions were easy to moderate,’ say teachers, students

According to Manendra Singh, TGT Hindi, Global Indian International School (GIIS), Noida, the Class 10 Hindi paper was a bit lengthy and the multiple choice questions were somewhat tricky.

trigonometry objective questions for class 10

The Central Board of Secondary Education (CBSE) conducted the Hindi course A and course B examinations of Class 10 on Wednesday. While experts said the papers were balanced, some students said they were a bit lengthy.

“The questions on Hindi course A grammar and creative writing were of a high standard,” said Dr Arvind Kumar of Uttar Pradesh Bulandshahr’s VidyaGyan School. The students solved the skill and value-based questions with intelligence, the Hindi PGT teacher said.

trigonometry objective questions for class 10

The questions, he added, from the supplementary book Kritika were skill and value-based. “All 17 questions were based on the syllabus, the options for the unseen prose and poetry were confusing. The students will score between 95% and 100%. The students took more time to solve the question paper,” Dr Kumar added.

The 80-mark Hindi course A question paper had two sections: section 1 comprised of multiple choice questions (40 marks) and section 2 consisted of descriptive questions (40 marks).

“The difficulty level of the CBSE 2024 Hindi Class 10 Course B questions was moderate,” said Shikha, TGT, KIIT World School, Gurugram. The Hindi question paper had a balanced mix of questions to check knowledge and analytical skills and were application-based. “Most of the students were able to complete the paper well in time,” she added.

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The Hindi course B exam was held for a total of 80 marks and was divided into sections A and B.

While section A had an unseen passage (10 marks) and grammar and literature MCQ (30 marks), section B had literature and a subjective part along with creative writing with internal choices (40 marks). “Both the sections in the paper were relatively easy,” said Shikha.

She added, “Section A passages were not tricky but the questions were value-based, while the second passage required an authority in the language.”

“The grammar section was simple with direct answers in sets 1 and 2. However, in set 2, a few questions were different from set 1,” said Shikha.

The TGT Hindi teacher further said, “Questions of paragraph writing and short story writing were easy. The literature part was easy too but it expected an examinee to have a thorough knowledge of the topics as well as subtopics with understanding of the themes, characters and literary techniques used in the text.”

“There was no error found in the question papers. Questions in Set 1, 2 and 3 were similar, but a few questions in the Literature section were different,” Shikha said while analysing the CBSE Class 10 Hindi paper.

“Overall, it was an easy-to-average paper as far as the difficulty level was concerned, with a balanced approach for all kinds of students. It was entirely based on the National Council of Educational Research and Training (NCERT) syllabus. The language was straightforward and a proper reading of questions would make it understandable for the students,” she added.

“Our students were able to tackle all the questions with ease and a positive attitude,” Singh added.

Saying that the questions in the writing, literature and grammar sections were straightforward and based on the CBSE syllabus, the teacher further said, “All the students were able to complete the paper on time and attempt all the questions with ease and accuracy as ample practice of sample papers had already been done by them.”

“Even the average students said that the paper was easy,” he added.

Saying that the Hindi question paper was a balanced one and framed as per the CBSE curriculum, Jyoti Tiwari, Head of the Department of Hindi, Billabong High International School, Malad ( Mumbai ) said: “Gadyansh questions were opinion based which was a good type of question for examinees. The unseen passage and the paragraph have been given in such a way that all the children can attempt it. Set 3 was a little tough. MCQs in Set 1 and 2 were a little challenging. Overall, a good question paper.”

According to students of Modern Public School, Shalimar Bagh and the Class 10 Hindi exam analysis shared by Alka Kapur, principal of the school, a few students found the test balanced, while others claimed that the passage section proved to be more challenging than the grammar one.

“The students felt that the three-hour time limit on the Hindi paper offered a substantial challenge in writing proficiency and required strong time management,” the principal said, adding that the majority of the literary portion’s questions were taken from the NCERT syllabus and were simple to understand.

As per Gayatri Sharma, educator, Senior Years, Shiv Nadar School Faridabad, students found the Hindi paper to be manageable.

“Unseen passages were consistent across all sets. Students noted that 2-3 questions were moderate, while the rest were quite good, with an average expected score of seven out of 10. In the grammar section, 70% of the questions were easy to attempt, but the remaining 30% were tricky. The similarity in the options for multiple-choice questions (MCQs) created some confusion among students,” Sharma added.

Avi Saraf, a student of the same school analysed the literature section as easy and lengthy. While questions from chapters and poems varied across sets, questions from ‘Sanchayan’ remained the same, he said.

“The creative writing section did not present significant challenges, and we could successfully express our thoughts in our own words,” the student added.

This time, as a first, the CBSE is allowing students suffering from Type-1 Diabetes to carry chocolates, fruits and snacks to the examination centres. “The students can carry the items in a transparent pouch,” the CBSE has said.

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  1. Trigonometry(objective type questions) Class:-10th

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  6. Term 1 Class 10th CBSE Trigonometry (Objective Questions)

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  1. MCQ Questions for Class 10 Maths Introduction to Trigonometry with

    1. The value of cos 0°. cos 1°. cos 2°. cos 3°… cos 89° cos 90° is (a) 1 (b) -1 (c) 0 (d) 1 2√ Answer 2. If x tan 45° sin 30° = cos 30° tan 30°, then x is equal to (a) √3 (b) 1 2 (c) 1 2√ (d) 1 Answer 3. If x and y are complementary angles, then (a) sin x = sin y (b) tan x = tan y (c) cos x = cos y (d) sec x = cosec y Answer

  2. Class 10 Maths Chapter 8 Introduction to Trigonometry MCQs

    Answer: (c) 1-√3 Explanation: sin 30° = ½, sin 60° = √3/2, cos 30° = √3/2 and cos 60° = ½ Putting these values, we get: (½+½)- (√3/2+√3/2) = 1 - [ (2√3)/2] = 1 - √3 3.

  3. CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry Objective

    Solution: In triangle ABC, ∠A + ∠B + ∠C =180 ° ∠A + ∠C= 180 ° - 90 ° = 90 ° ⇒ None of the angles can be ≥ 90 ° In a right triangle ABC, the right angle is at B. What is the length of the missing side in the figure? 25 cm 12 cm 7 cm 5 cm Answer: (D) 5 cm Solution: Pythagoras theorem: In a right-angled triangle,

  4. MCQ Questions for Class 10 Maths with Answers Chapter 8 Trigonometry

    MCQ Questions for Class 10 Maths Chapter 8 Trigonometry. Q.2. In a right triangle ABC, the right angle is at B. Which of the following is true about the other two angles A and C? Q.3. (cos A / cot A) + sin A= ? Q.4. In a right triangle ABC, the right angle is at B. What is the length of missing side in the figure? Q.5.

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    Question 1. If tan θ + cot θ = 5, find the value of tan2θ + cotθ. (2012) Solution: tan θ + cot θ = 5 … [Given tan 2 θ + cot 2 θ + 2 tan θ cot θ = 25 … [Squaring both sides tan 2 θ + cot 2 θ + 2 = 25 ∴ tan 2 θ + cot 2 θ = 23 Question 2. If sec 2A = cosec (A - 27°) where 2A is an acute angle, find the measure of ∠A. (2012, 2017D) Solution:

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    Answer/ Explanation 2. In given figure, the value of CE is (a) 12 cm (b) 6 cm (c) 9 cm (d) 6√3 cm Answer/ Explanation MCQ Questions For Class 10 Maths Trigonometry Question 3. In given figure, the value of ZC is (a) 90° (b) 45° (c) 30° (d) 60° Answer/ Explanation MCQ On Application Of Trigonometry Question 4.

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    Question. 1 : In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C Solution: In a given triangle ABC, right-angled at B = ∠B = 90° Given: AB = 24 cm and BC = 7 cm That means, AC = Hypotenuse According to the Pythagoras Theorem,

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    Trigonometry Multiple Choice Questions And Answers Pdf Question 13. The upper part of a tree is broken by the wind and makes an angle of 30° with the ground. ... Trigonometry Objective Questions For Class 10 Question 24. The angle of elevation of the top of a 15m high tower at a point 15m away from the base of the tower is ____ . Answer ...

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  14. Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8

    Introduction to Trigonometry Class 10 Extra Questions Short Answer Type 1. Question 1. Find maximum value of 1 sec θ, 0°≤ θ ≤ 90°. Solution: 1 sec θ, (0° ≤ θ ≤ 90°) (Given) ∵ sec θ is in the denominator. ∴ The min. value of sec θ will return max. value for 1sec θ. But the min. value of sec θ is sec 0° = 1.

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    Class 10 Maths MCQs Chapter 8 Introduction to Trigonometry. 1. The value of cos 0°. cos 1°. cos 2°. cos 3°… cos 89° cos 90° is. 2. If x tan 45° sin 30° = cos 30° tan 30°, then x is equal to. 3. If x and y are complementary angles, then. 5.

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    Very Short Answer Type Questions [1 Mark] Question 1. Find the value of sec² 42° - cosec² 48°. Solution: Question 2. If (1 + cos A) (1 - cos A) =3/4, find the value of sec A. Solution: Question 3. If cosec θ + cot θ = x, find the value of cosec θ - cot θ.

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    Introduction to Trigonometry Class 10 Extra Questions HOTS. Question 1. Show that cosec 2 θ + sec 2 θ can never be less than 2. Answer: Let, if possible, cosec 2 θ + sec 2 θ < 2. ⇒ (1 + cot 2 θ) + (1 + tan 2 θ) < 2. ⇒ cot 2 θ + tan 2 θ + 2 < 2. ⇒ cot 2 θ + tan 2 θ < θ. But sum of squares of two real numbers is always non negative.

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    Solution: Ex 8.1 Class 10 Maths Question 4. Given 15 cot A = 8, find sin A and sec A. Solution: Ex 8.1 Class 10 Maths Question 5. Given sec θ = 1312 , calculate all other trigonometric ratios. Solution: Ex 8.1 Class 10 Maths Question 6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

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  20. Class 10 Maths Chapter 18 Trigonometry Important Questions

    Ans: Trigonometry is the study of the relationships between the angles and sides of triangles. It primarily focuses on right triangles, which have one angle equal to 90 degrees. Trigonometric functions and ratios, such as sine, cosine, and tangent, are used to establish these relationships. Key Concepts and Objectives:

  21. Class 10 Maths Chapter 9 Some Applications of Trigonometry MCQs

    Answer: (a) Increasing Explanation: See the following figure: As the shadow reaches from point D to C towards the direction of the tree, the angle of elevation increase from 30° to 60°. 2. The angle of elevation of the top of a building from a point on the ground, which is 30 m away from the foot of the building, is 30°.

  22. CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Objective

    CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Objective Questions deal with how trigonometry helps to find the height and distance of different objects without any measurement. In earlier days, astronomers used trigonometry to calculate the distance from the earth to the planets and stars.

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  24. Class 10 Math MCQ PDF (Questions and Answers)

    The Book Class 10 Math Multiple Choice Questions (MCQ Quiz) with Answers PDF Download (10th Grade Math Book): MCQ Questions Chapter 1-13 & Practice Tests with Answers Key (Grade 10 Math Textbook MCQs, Notes & Study Guide) covers review tests for competitive exams with solved MCQs. …

  25. CBSE Class 10 Hindi exam analysis: 'Questions were easy to moderate

    According to Manendra Singh, TGT Hindi, Global Indian International School (GIIS), Noida, the Class 10 Hindi paper was a bit lengthy and the multiple choice questions were somewhat tricky. "Our students were able to tackle all the questions with ease and a positive attitude," Singh added.