# NCERT Solutions For Class 10 Maths Chapter 8 PDF - Introduction to Trigonometry

## NCERT Solutions For Class 10 Maths Chapter 8 PDF Free Download

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry is helpful for the students as it helps to understand the concepts and score well in board examination. The solutions are designed and reviewed by the subject experts that covers all the questions from the textbook. These NCERT solutions are framed as per the latest NCERT syllabus and guidelines and also the exam pattern of the CBSE board.

The solutions provide a strong foundation for every concept. Students can clarify their doubts and understand the fundamentals present in the chapter. They should be able to solve the difficult problems in each exercise with the help of NCERT Solutions for Class 10.

### Class 10 Maths Chapter 8 Exercise 8.1 Page: 181

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

According to the Pythagoras Theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get

AC2=AB2+BC2

AC2 (24)2+72

AC2 =(576+49)

AC2 = 625cm2

Therefore, AC = 25 cm

2. In Fig. 8.13, find tan P – cot R

3. If sin A = 3/4 , Calculate cos A and tan A.

Solution:

4. Given 15 cot A = 8, find sin A and sec A.

Solution:

5. Given sec θ = 13/12 Calculate all other trigonometric ratios

Solution:

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Solution:

Let us assume the triangle ABC in which CD⊥AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

Now, interchange the terms, we get

Let take a constant value

Now consider the equation as

AC = K BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD2 = BC2 – BD2 … (3)

From the equations (3) and (4) we get,

Now substitute the equations (1) and (2) in (3) and (4)

K2(BC2−BD2)=(BC2−BD2) k2=1

Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

7. If cot θ = 7/8, evaluate :

(i) (1+sinθ)(1-sinθ) / (1+cosθ)(1-cosθ)

(ii) cot2 θ

Solution:

Let us assume a △ABC in which  ∠B=90° and ∠C= θ

8. If 3 cot A = 4, check whether $\frac{1-tan^{2}A}{1+tan^{2}A}$ = cos2 A – sin 2 A or not.

9. In triangle ABC, right-angled at B, if tan A =1/√ 3.Find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii)cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 4/3 for some angle θ

### Exercise 8.2 Page 187

2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan230° =
(A) sin 60°            (B) cos 60°          (C) tan 60°            (D) sin 30°
(ii) 1-tan245°/1+tan245° =
(A) tan 90°            (B) 1                    (C) sin 45°            (D) 0
(iii)  sin 2A = 2 sin A is true when A =
(A) 0°                   (B) 30°                  (C) 45°                 (D) 60°

(iv) 2tan30°/1-tan230° =
(A) cos 60°          (B) sin 60°             (C) tan 60°           (D) sin 30°

3. If tan (A + B) =√3 and tan (A – B) =1/√3,0° < A + B ≤ 90°; A > B, find A and B.

Solution:

tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

Now substitute the degree value

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A= 45°

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

4. State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

### Class 10 Maths Chapter 8 Exercise 8.3 Page: 189

1. Evaluate :

(i) sin 18°/cos 72°

(ii) tan 26°/cot 64°

(iii)  cos 48° – sin 42°

(iv)  cosec 31° – sec 59°

Solution:

(i) sin 18°/cos 72°

To simplify this, convert the sin function into cos function

We know that, 18° is written as 90° – 18°, which is equal to the cos 72°.

= sin (90° – 18°) /cos 72°

Substitute the value, to simplify this equation

= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

To simplify this, convert the tan function into cot function

We know that, 26° is written as 90° – 36°, which is equal to the cot 64°.

= tan (90° – 36°)/cot 64°

Substitute the value, to simplify this equation

= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°

To simplify this, convert the cos function into sin function

We know that, 48° is written as 90° – 42°, which is equal to the sin 42°.

= cos (90° – 42°) – sin 42°

Substitute the value, to simplify this equation

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

To simplify this, convert the cosec function into sec function

We know that, 31° is written as 90° – 59°, which is equal to the sec 59°.

= cosec (90° – 59°) – sec 59°

Substitute the value, to simplify this equation

= sec 59° – sec 59° = 0

2.  Show that :

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution:

(i) tan 48° tan 23° tan 42° tan 67°

Simplify the given problem by converting some of the tan functions to the cot functions

We know that tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute the values

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°

Simplify the given problem by converting some of the cos functions to the sin functions

We know that cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°

Substitute the values

= sin 52° sin 38° – sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

Therefore, the value of A = 36°

4.  If tan A = cot B, prove that A + B = 90°.

Solution:

tan A = cot B

We know that cot B = tan (90° – B)

To prove A + B = 90°, substitute the above equation in the given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

Hence Proved.

5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution:

sec 4A = cosec (A – 20°)

We know that sec 4A = cosec (90° – 4A)

To find the value of A , substitute the above equation in the given problem

cosec (90° – 4A) = cosec (A – 20°)

Now, equate the angles

90° – 4A= A- 20°

110° = 5A

A = 110°/ 5 = 22°

Therefore, the value of A = 22°

6. If A, B and C are interior angles of a triangle ABC, then show that

sin (B+C/2) = cos A/2

Solution:

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

Since sin (90°-A/2) = = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

Hence proved.

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

Given:

sin 67° + cos 75°

In term of sin as cos function and cos as sin function, it can be written as follows

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

= sin (90° – 23°) + cos (90° – 15°)

Now, simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°

### Class 10 Maths Chapter 8 Exercise 8.4 Page: 193

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution:

3. Evaluate :
(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii)  sin 25° cos 65° + cos 25° sin 65°

Solution:

(i) (sin263° + sin227°)/(cos217° + cos273°)

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= [sin2(90°-27°) + sin227°] / [cos2(90°-73°) + cos273°)]

= (cos227° + sin227°)/(sin227° + cos273°)

= 1/1 =1                       (since sin2A + cos2A = 1)

Therefore, (sin263° + sin227°)/(cos217° + cos273°) = 1

(ii) sin 25° cos 65° + cos 25° sin 65°

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos265° + sin265° = 1 (since sin2A + cos2A = 1)

Therefore, sin 25° cos 65° + cos 25° sin 65° = 1

4. Choose the correct option. Justify your choice.
(i) 9 sec2A – 9 tan2A =
(A) 1                 (B) 9              (C) 8                (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0                 (B) 1              (C) 2                (D) – 1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A           (B) sin A        (C) cosec A      (D) cos A

(iv) 1+tan2A/1+cot2A =

(A) sec2 A                 (B) -1              (C) cot2A                (D) tan2A

(iv) (D) is correct.

Justification:

We know that,

tan2A =1/cot2A

Now, substitute this in the given problem, we get

1+tan2A/1+cot2A

= (1+1/cot2A)/1+cot2A

= (cot2A+1/cot2A)×(1/1+cot2A)

= 1/cot2A = tan2A

So, 1+tan2A/1+cot2A = tan2A

5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.

(i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

[Hint : Write the expression in terms of sin θ and cos θ]

(iv) (1 + sec A)/sec A = sin2A/(1-cos A)

[Hint : Simplify LHS and RHS separately]

(v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec2A = 1+cot2A.

(vi) √[1+sinA/1-sinA] = sec A + tan A

(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A

## NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry.

For class 10 CBSE Maths paper, out of the 80 marks, 12 marks are assigned from the unit 5 “Trigonometry”. The paper consists of 4 parts. Each part carries different marks and the questions have been assigned with 1 mark, two marks, 3 marks and 4 marks. You can expect at least 2-3 compulsory questions from this chapter. The main topics covered in this chapter include:

8.1 Introduction

You have already studied about triangles, and in particular, right triangles, in your earlier classes. In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We also define the trigonometric ratios for angles of measure 0o and 90o. We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

8.2 Trigonometric Identities

You have studied the concept of ratio, in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios. The topic is explained with suitable examples by using different functions of Trigonometry.

8.3 Trigonometric Ratios of Some Specific Angles

From Geometry, you are already familiar with the construction of angles of 300, 45o, 60o and 900. In this section, we will find the values of the trigonometric ratios for these angles and for 0o. It explains Trigonometric Ratios of 45o, Trigonometric Ratios of 30o and 600, Trigonometric Ratios of 0o and 90o with suitable examples.

8.4 Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum equals 90o. The topic discusses various formulas to solve numerical problems related to trigonometric ratios.

8.5 Trigonometric Identities

You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true of all values of the angles involved. In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.

8.6 Summary

Summary is the brief of concepts which includes all the important points which you need to memorize to solve numerical problems related to the chapter.

List of Exercises in class 10 Maths Chapter 8 :

Exercise 8.1 Solutions – 11 Questions (7 short answers, 3 long answers, 1 short answer with reasoning)

Exercise 8.2 Solutions – 4 Questions ( 1 short answer, 2 long answers, 1 MCQ)

Exercise 8.3 Solutions – 7 Questions (5 short answers, 2 long answers)

Exercise 8.4 Solutions – 5 Questions ( 2 short answers, 2 long answers, 1 MCQ)

In this chapter, class 10 students are introduced with the concept of trigonometry as it involves the study of some important ratios of the right angle triangle with respect to the angles called trigonometric ratios of the angles. Students will also learn to establish some of the trigonometric identities and solve some specific angles(from 0° to 90°.) of the trigonometric ratios, sine, cosine, tangents etc.

Learn concepts like ratios of complementary angles, specific angles from the NCERT class 10 textbooks. Here you are provided with the solutions of all the questions for 10th maths chapter that covers in the given topics.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry

• Solutions are given in step by step procedure
• Helps to increase the efficiency of solving the questions
• The solutions are properly organised, chapter-wise, which students can find easily
• Designed by subjects experts to deliver accurate answers.
• A valuable aid to the students for the preparation of board examinations.

 Also Access NCERT Exemplar for Class 10 Maths Chapter 8 Introduction To Trigonometry CBSE Notes for Class 10 Maths Chapter 8 Introduction To Trigonometry

## Frequently Asked Questions on Chapter 8 – Introduction to Trigonometry

### If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B?

Let us assume the triangle ABC in which CD⊥AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

Now, interchange the terms, we get

Let take a constant value

Now consider the equation as

AC = K BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD2=BC2-BD2…(3)

From the equations (3) and (4) we get,

Now substitute the equations (1) and (2) in (3) and (4)

k2(BC-BD2)=(BC2-BD2)K2=1

Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

### If tan (A + B) =√3 and tan (A – B) =1/√3,0° < A + B ≤ 90°; A > B, find A and B?

tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

Now substitute the degree value

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A= 45°

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

### Evaluate: sin 18°/cos 72°?

(i) sin 18°/cos 72°

To simplify this, convert the sin function into cos function

We know that, 18° is written as 90° – 18°, which is equal to the cos 72°.

= sin (90° – 18°) /cos 72°

Substitute the value, to simplify this equation

= cos 72° /cos 72° = 1

### Show that : tan 48° tan 23° tan 42° tan 67° = 1?

tan 48° tan 23° tan 42° tan 67°

Simplify the given problem by converting some of the tan functions to the cot functions

We know that tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute the values

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

### If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A?

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

Therefore, the value of A = 36°