---
title: "NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3"
url: https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-3
dateModified: 2026-10-07T15:53:32+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3

Chapter 8: Introduction to Trigonometry. Every question from Exercise 8.3, with full working and the final answer.

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## EXERCISE 8.3

### Question 1

*3 marks · Proof*

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

**Solution**

1. We know that $\tan A = \frac{1}{\cot A}$.
2. Also, $\sin A = \frac{1}{\csc A}$, and since $\csc^2 A = 1 + \cot^2 A$, we have $\csc A = \sqrt{1 + \cot^2 A}$.
3. Therefore, $\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$.
4. For $\sec A$, we use $\sec^2 A = 1 + \tan^2 A = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A}$.
5. Taking the square root, we get $\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot^2 A}$ (or $\frac{\sqrt{1 + \cot^2 A}}{\cot A}$ since angles are acute).
6. Hence expressed.

**Answer:** $\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$, $\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$, $\tan A = \frac{1}{\cot A}$

> Common mistake: Forgetting to take the square root properly when converting secant and sine.

### Question 2

*4 marks · Proof*

Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

**Solution**

1. We know that $\cos A = \frac{1}{\sec A}$.
2. Using $\sin^2 A + \cos^2 A = 1$, we get $\sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A}$, so $\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$.
3. For $\csc A$, taking the reciprocal of $\sin A$ gives $\csc A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$.
4. For $\tan A$, using $\tan^2 A = \sec^2 A - 1$, we get $\tan A = \sqrt{\sec^2 A - 1}$.
5. For $\cot A$, taking the reciprocal of $\tan A$ gives $\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$.
6. Hence all ratios are expressed in terms of $\sec A$.

**Answer:** $\cos A = \frac{1}{\sec A}$, $\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$, $\csc A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$, $\tan A = \sqrt{\sec^2 A - 1}$, $\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$

> Common mistake: Mixing up the reciprocals and square roots.

### Question 3

*1 mark · MCQ*

Choose the correct option. Justify your choice.
(i) $9 \sec^2 A - 9 \tan^2 A =$
(ii) $(1 + \tan \theta + \sec \theta)(1 + \cot \theta - \csc \theta) =$
(iii) $(\sec A + \tan A)(1 - \sin A) =$
(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$

- 1, 9, 8, 0
- 0, 1, 2, -1
- \sec A, \sin A, \csc A, \cos A
- \sec^2 A, -1, \cot^2 A, \tan^2 A

**Part (i)**

1. $9 \sec^2 A - 9 \tan^2 A = 9(\sec^2 A - \tan^2 A)$
2. Using the identity $\sec^2 A - \tan^2 A = 1$, we get $9(1) = 9$.

Answer (i): (B) 9

**Part (ii)**

1. Expressing in terms of sine and cosine: $\left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right)\left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)$
2. Simplifying each bracket: $\left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right)\left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right)$
3. Using $(a+b)(a-b)$ in the numerator: $\frac{(\cos \theta + \sin \theta)^2 - 1^2}{\sin \theta \cos \theta}$
4. Expanding and simplifying: $\frac{\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta} = \frac{1 + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta} = 2$

Answer (ii): (C) 2

**Part (iii)**

1. Converting to sine and cosine: $\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 - \sin A)$
2. Combining terms: $\left(\frac{1 + \sin A}{\cos A}\right)(1 - \sin A) = \frac{1 - \sin^2 A}{\cos A}$
3. Since $1 - \sin^2 A = \cos^2 A$, we get $\frac{\cos^2 A}{\cos A} = \cos A$.

Answer (iii): (D) \cos A

**Part (iv)**

1. Writing in terms of sine and cosine: $\frac{1 + \frac{\sin^2 A}{\cos^2 A}}{1 + \frac{\cos^2 A}{\sin^2 A}}$
2. Simplifying the numerator and denominator: $\frac{\frac{\cos^2 A + \sin^2 A}{\cos^2 A}}{\frac{\sin^2 A + \cos^2 A}{\sin^2 A}} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}$
3. Reciprocating and multiplying: $\frac{\sin^2 A}{\cos^2 A} = \tan^2 A$

Answer (iv): (D) \tan^2 A

**Answer:** (i) (B), (ii) (B), (iii) (D), (iv) (D)

> Common mistake: Forgetting standard trigonometric identities like $\sec^2 A - \tan^2 A = 1$ or making algebraic expansion errors in part (ii).

### Question 4

*5 marks · Proof*

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) $(\csc \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
(ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
(iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$
(iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}$
(v) $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$
(vi) $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$
(vii) $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
(viii) $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
(ix) $(\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$
(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

**Part (i)**

1. LHS $= (\csc \theta - \cot \theta)^2 = \left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$.
2. Substitute $\sin^2 \theta = 1 - \cos^2 \theta$: $\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)}$.
3. Cancel $(1 - \cos \theta)$ to get $\frac{1 - \cos \theta}{1 + \cos \theta} =$ RHS.
4. Hence proved.

Answer (i): Hence proved.

**Part (ii)**

1. LHS $= \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A}$.
2. Expand numerator: $\frac{\cos^2 A + 1 + \sin^2 A + 2 \sin A}{(1 + \sin A)\cos A} = \frac{(\cos^2 A + \sin^2 A) + 1 + 2 \sin A}{(1 + \sin A)\cos A}$.
3. Since $\cos^2 A + \sin^2 A = 1$, we get $\frac{1 + 1 + 2 \sin A}{(1 + \sin A)\cos A} = \frac{2 + 2 \sin A}{(1 + \sin A)\cos A} = \frac{2(1 + \sin A)}{(1 + \sin A)\cos A}$.
4. Cancel $(1 + \sin A)$ to get $\frac{2}{\cos A} = 2 \sec A =$ RHS.
5. Hence proved.

Answer (ii): Hence proved.

**Part (iii)**

1. Write in terms of sine and cosine: $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$.
2. Simplify fractions: $\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)} = \frac{1}{\sin \theta - \cos \theta} \left(\frac{\sin^3 \theta - \cos^3 \theta}{\cos \theta \sin \theta}\right)$.
3. Use algebraic formula for difference of cubes: $\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\cos \theta \sin \theta (\sin \theta - \cos \theta)} = \frac{1 + \sin \theta \cos \theta}{\cos \theta \sin \theta}$.
4. Split terms: $\frac{1}{\cos \theta \sin \theta} + \frac{\sin \theta \cos \theta}{\cos \theta \sin \theta} = \sec \theta \csc \theta + 1 =$ RHS.
5. Hence proved.

Answer (iii): Hence proved.

**Part (iv)**

1. LHS $= \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = 1 + \cos A$.
2. Multiply numerator and denominator by $(1 - \cos A)$: $\frac{(1 + \cos A)(1 - \cos A)}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{\sin^2 A}{1 - \cos A} =$ RHS.
3. Hence proved.

Answer (iv): Hence proved.

**Part (v)**

1. Divide numerator and denominator of LHS by $\sin A$: $\frac{\frac{\cos A}{\sin A} - 1 + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + 1 - \frac{1}{\sin A}} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A}$.
2. Rearrange terms using $\csc^2 A - \cot^2 A = 1$: $\frac{(\csc A + \cot A) - (\csc^2 A - \cot^2 A)}{\cot A + 1 - \csc A} = \frac{(\csc A + \cot A)[1 - (\csc A - \cot A)]}{\cot A + 1 - \csc A}$.
3. Simplify and cancel common bracket: $\frac{(\csc A + \cot A)(1 - \csc A + \cot A)}{1 - \csc A + \cot A} = \csc A + \cot A =$ RHS.
4. Hence proved.

Answer (v): Hence proved.

**Part (vi)**

1. Rationalize inside the square root: $\sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$.
2. Substitute $1 - \sin^2 A = \cos^2 A$: $\sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A}$.
3. Split terms: $\frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A =$ RHS.
4. Hence proved.

Answer (vi): Hence proved.

**Part (vii)**

1. Take out common factors: LHS $= \frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2 \cos^2 \theta - 1)}$.
2. Substitute $\cos^2 \theta = 1 - \sin^2 \theta$ in the denominator: $\frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2(1 - \sin^2 \theta) - 1)} = \frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2 - 2 \sin^2 \theta - 1)}$.
3. Simplify denominator: $\frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(1 - 2 \sin^2 \theta)} = \frac{\sin \theta}{\cos \theta} = \tan \theta =$ RHS.
4. Hence proved.

Answer (vii): Hence proved.

**Part (viii)**

1. Expand both squares: LHS $= (\sin^2 A + \csc^2 A + 2 \sin A \csc A) + (\cos^2 A + \sec^2 A + 2 \cos A \sec A)$.
2. Group sine and cosine terms: $(\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 2(1) + 2(1)$.
3. Substitute identities $\sin^2 A + \cos^2 A = 1$, $\csc^2 A = 1 + \cot^2 A$, and $\sec^2 A = 1 + \tan^2 A$: $1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 4$.
4. Combine constants: $1 + 1 + 1 + 4 + \tan^2 A + \cot^2 A = 7 + \tan^2 A + \cot^2 A =$ RHS.
5. Hence proved.

Answer (viii): Hence proved.

**Part (ix)**

1. Simplify LHS: $\left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) = \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right) = \frac{\cos^2 A \sin^2 A}{\sin A \cos A} = \sin A \cos A$.
2. Simplify RHS: $\frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\cos A \sin A}} = \frac{1}{\frac{1}{\cos A \sin A}} = \cos A \sin A$.
3. Since LHS = RHS.
4. Hence proved.

Answer (ix): Hence proved.

**Part (x)**

1. Simplify the first term: $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$.
2. Simplify the second term: $\left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2 = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2 = (-\tan A)^2 = \tan^2 A$.
3. Since all parts equal $\tan^2 A$.
4. Hence proved.

Answer (x): Hence proved.

**Answer:** All identities proved successfully.

> Common mistake: Algebraic expansion errors or incorrect application of square identities.

## Related pages

- [All Chapter 8 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions)
- [Exercise 8.1](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-1)
- [Exercise 8.2](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-2)

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