NCERT Solutions
Class 10 - Mathematics - Chapter 8: INTRODUCTION TO TRIGONOMETRY

Exercise 8.3

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

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

Choose the correct option. Justify your choice:

1. \(9 \sec^2 A - 9 \tan^2 A =\)
   (A) 1   (B) 9   (C) 8   (D) 0
2. \((1 + \tan \theta + \sec \theta)(1 + \cot \theta - \text{cosec } \theta) =\)
   (A) 0   (B) 1   (C) 2   (D) -1
3. \((\sec A + \tan A)(1 - \sin A) =\)
   (A) \sec A   (B) \sin A   (C) \text{cosec } A   (D) \cos A
4. \(\dfrac{1 + \tan^2 A}{1 + \cot^2 A} =\)
   (A) \sec^2 A   (B) -1   (C) \cot^2 A   (D) \tan^2 A

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

1. \((\text{cosec } \theta - \cot \theta)^2 = \dfrac{1 - \cos \theta}{1 + \cos \theta}\)
2. \(\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2 \sec A\)
3. \(\dfrac{\tan \theta}{1 - \cot \theta} + \dfrac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta\)
4. \(\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A}\)
5. \(\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A\)
6. \(\sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A\)
7. \(\dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta\)
8. \((\sin A + \text{cosec} A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A\)
9. \((\text{cosec } A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A}\)
10. \(\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \left( \dfrac{1 - \tan A}{1 - \cot A} \right)^2 = \tan^2 A\)