NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3#
Exercise 8.3 covers Trigonometric Identities. Students express trigonometric ratios in terms of other ratios and evaluate expressions using fundamental identities.
Fundamental Identities:
- $\sin^2 A + \cos^2 A = 1$
- $1 + \tan^2 A = \sec^2 A$
- $1 + \cot^2 A = \cosec^2 A$
1. Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$. #
Answer
sin A in terms of cot A:
Using $1 + \cot^2 A = \cosec^2 A$:
$$\cosec^2 A = 1 + \cot^2 A$$$$\Rightarrow \cosec A = \sqrt{1 + \cot^2 A}$$
$$\Rightarrow \sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$$
tan A in terms of cot A:
$$\tan A = \frac{1}{\cot A}$$sec A in terms of cot A:
Using $1 + \tan^2 A = \sec^2 A$:
$$\sec^2 A = 1 + \frac{1}{\cot^2 A}$$$$\Rightarrow \sec^2 A = \frac{\cot^2 A + 1}{\cot^2 A}$$
$$\Rightarrow \sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$$
∴ $\sin A = \mathbf{\dfrac{1}{\sqrt{1 + \cot^2 A}}}$, $\sec A = \mathbf{\dfrac{\sqrt{1 + \cot^2 A}}{\cot A}}$ and $\tan A = \mathbf{\dfrac{1}{\cot A}}$
2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$. #
Answer
cos A:
$$\cos A = \frac{1}{\sec A}$$tan A:
Using $1 + \tan^2 A = \sec^2 A$:
$$\tan^2 A = \sec^2 A - 1$$$$\Rightarrow \tan A = \sqrt{\sec^2 A - 1}$$
sin A:
Using $\sin^2 A + \cos^2 A = 1$:
$$\sin^2 A = 1 - \frac{1}{\sec^2 A}$$$$\Rightarrow \sin^2 A = \frac{\sec^2 A - 1}{\sec^2 A}$$
$$\Rightarrow \sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$$
cot A:
$$\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}$$cosec A:
$$\cosec A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$$3. Choose the correct option. Justify your choice. #
(i) $9\sec^2 A - 9\tan^2 A =$ (A) 1 (B) 9 (C) 8 (D) 0
We simplify $9\sec^2 A - 9\tan^2 A$:
$$= 9(\sec^2 A - \tan^2 A)$$$$= 9 \times 1$$
$$= 9$$
(Using $\sec^2 A - \tan^2 A = 1$.)
∴ Answer: (B) 9
(ii) $(1 + \tan\theta + \sec\theta)(1 + \cot\theta - \cosec\theta) =$ (A) 0 (B) 1 (C) 2 (D) –1
We simplify $(1 + \tan\theta + \sec\theta)(1 + \cot\theta - \cosec\theta)$:
$$= \left(1 + \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta}\right)\left(1 + \frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta}\right)$$$$= \frac{\cos\theta + \sin\theta + 1}{\cos\theta} \times \frac{\sin\theta + \cos\theta - 1}{\sin\theta}$$
$$= \frac{(\sin\theta + \cos\theta)^2 - 1}{\sin\theta\cos\theta}$$
$$= \frac{\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta - 1}{\sin\theta\cos\theta}$$
$$= \frac{1 + 2\sin\theta\cos\theta - 1}{\sin\theta\cos\theta}$$
$$= 2$$
∴ Answer: (C) 2
(iii) $(\sec A + \tan A)(1 - \sin A) =$ (A) sec A (B) sin A (C) cosec A (D) cos A
We simplify $(\sec A + \tan A)(1 - \sin A)$:
$$= \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 - \sin A)$$$$= \frac{(1 + \sin A)(1 - \sin A)}{\cos A}$$
$$= \frac{1 - \sin^2 A}{\cos A}$$
$$= \frac{\cos^2 A}{\cos A}$$
$$= \cos A$$
∴ Answer: (D) cos A
(iv) $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} =$ (A) sec²A (B) –1 (C) cot²A (D) tan²A
We simplify $\dfrac{1 + \tan^2 A}{1 + \cot^2 A}$:
$$= \frac{\sec^2 A}{\cosec^2 A}$$$$= \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}$$
$$= \frac{\sin^2 A}{\cos^2 A}$$
$$= \tan^2 A$$
∴ Answer: (D) tan²A
4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. #
(i) $(\cosec\theta - \cot\theta)^2 = \dfrac{1 - \cos\theta}{1 + \cos\theta}$
Taking the LHS, $(\cosec\theta - \cot\theta)^2$:
$$= \left(\frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta}\right)^2$$$$= \frac{(1 - \cos\theta)^2}{\sin^2\theta}$$
$$= \frac{(1 - \cos\theta)^2}{1 - \cos^2\theta}$$
$$= \frac{(1 - \cos\theta)^2}{(1 - \cos\theta)(1 + \cos\theta)}$$
$$= \frac{1 - \cos\theta}{1 + \cos\theta}$$
∴ LHS = RHS. Hence proved. $\blacksquare$
(ii) $\dfrac{\cos A}{1+\sin A} + \dfrac{1+\sin A}{\cos A} = 2\sec A$
Taking the LHS, $\dfrac{\cos A}{1+\sin A} + \dfrac{1+\sin A}{\cos A}$:
$$= \frac{\cos^2 A + (1 + \sin A)^2}{\cos A(1 + \sin A)}$$$$= \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{\cos A(1 + \sin A)}$$
$$= \frac{2 + 2\sin A}{\cos A(1 + \sin A)}$$
$$= \frac{2(1 + \sin A)}{\cos A(1 + \sin A)}$$
$$= \frac{2}{\cos A}$$
$$= 2\sec A$$
∴ LHS = RHS. Hence proved. $\blacksquare$
(iii) $\dfrac{\tan\theta}{1-\cot\theta} + \dfrac{\cot\theta}{1-\tan\theta} = 1 + \sec\theta\cosec\theta$ [Hint: Write the expression in terms of $\sin\theta$ and $\cos\theta$]
Taking the LHS, $\dfrac{\tan\theta}{1-\cot\theta} + \dfrac{\cot\theta}{1-\tan\theta}$:
$$= \frac{\frac{\sin\theta}{\cos\theta}}{\frac{\sin\theta - \cos\theta}{\sin\theta}} + \frac{\frac{\cos\theta}{\sin\theta}}{\frac{\cos\theta - \sin\theta}{\cos\theta}}$$$$= \frac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} - \frac{\cos^2\theta}{\sin\theta(\sin\theta - \cos\theta)}$$
$$= \frac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)}$$
$$= \frac{(\sin\theta - \cos\theta)(\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta)}{\sin\theta\cos\theta(\sin\theta - \cos\theta)}$$
$$= \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta}$$
$$= \frac{1}{\sin\theta\cos\theta} + 1$$
$$= 1 + \sec\theta\cosec\theta$$
∴ LHS = RHS. Hence proved. $\blacksquare$
(iv) $\dfrac{1+\sec A}{\sec A} = \dfrac{\sin^2 A}{1-\cos A}$ [Hint: Simplify LHS and RHS separately]
Taking the LHS, $\dfrac{1+\sec A}{\sec A}$:
$$= \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}$$$$= \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}}$$
$$= 1 + \cos A$$
Taking the RHS, $\dfrac{\sin^2 A}{1-\cos A}$:
$$= \frac{1 - \cos^2 A}{1 - \cos A}$$$$= \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A}$$
$$= 1 + \cos A$$
∴ LHS = RHS. Hence proved. $\blacksquare$
(v) $\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \cosec A + \cot A$, using the identity $\cosec^2 A = 1 + \cot^2 A$
Taking the LHS and dividing the numerator and denominator by $\sin A$:
$$= \frac{\cot A - 1 + \cosec A}{\cot A + 1 - \cosec A}$$Using $\cosec^2 A - \cot^2 A = 1$ to replace the 1 in the numerator:
$$= \frac{(\cot A + \cosec A) - (\cosec^2 A - \cot^2 A)}{\cot A + 1 - \cosec A}$$$$= \frac{(\cot A + \cosec A) - (\cosec A + \cot A)(\cosec A - \cot A)}{\cot A + 1 - \cosec A}$$
$$= \frac{(\cot A + \cosec A)(1 - \cosec A + \cot A)}{\cot A + 1 - \cosec A}$$
$$= \cosec A + \cot A$$
∴ LHS = RHS. Hence proved. $\blacksquare$
(vi) $\sqrt{\dfrac{1+\sin A}{1-\sin A}} = \sec A + \tan A$
Taking the LHS and multiplying the numerator and denominator inside the root by $(1 + \sin A)$:
$$= \sqrt{\frac{(1 + \sin A)^2}{(1 - \sin A)(1 + \sin A)}}$$$$= \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$$
$$= \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$$
$$= \frac{1 + \sin A}{\cos A}$$
$$= \frac{1}{\cos A} + \frac{\sin A}{\cos A}$$
$$= \sec A + \tan A$$
∴ LHS = RHS. Hence proved. $\blacksquare$
(vii) $\dfrac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \tan\theta$
Taking the LHS, $\dfrac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta}$:
$$= \frac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(2\cos^2\theta - 1)}$$$$= \frac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta[2(1 - \sin^2\theta) - 1]}$$
$$= \frac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(1 - 2\sin^2\theta)}$$
$$= \frac{\sin\theta}{\cos\theta}$$
$$= \tan\theta$$
∴ LHS = RHS. Hence proved. $\blacksquare$
(viii) $(\sin A + \cosec A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
Taking the LHS, $(\sin A + \cosec A)^2 + (\cos A + \sec A)^2$:
$$= \sin^2 A + 2\sin A\cosec A + \cosec^2 A + \cos^2 A + 2\cos A\sec A + \sec^2 A$$$$= (\sin^2 A + \cos^2 A) + 2 + 2 + \cosec^2 A + \sec^2 A$$
$$= 1 + 4 + (1 + \cot^2 A) + (1 + \tan^2 A)$$
$$= 7 + \tan^2 A + \cot^2 A$$
(Using $\sin A \cosec A = 1$ and $\cos A \sec A = 1$.)
∴ LHS = RHS. Hence proved. $\blacksquare$
(ix) $(\cosec A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A}$ [Hint: Simplify LHS and RHS separately]
Taking the LHS, $(\cosec A - \sin A)(\sec A - \cos A)$:
$$= \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right)$$$$= \frac{1 - \sin^2 A}{\sin A} \times \frac{1 - \cos^2 A}{\cos A}$$
$$= \frac{\cos^2 A}{\sin A} \times \frac{\sin^2 A}{\cos A}$$
$$= \sin A\cos A$$
Taking the RHS, $\dfrac{1}{\tan A + \cot A}$:
$$= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$$$$= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A\cos A}}$$
$$= \sin A\cos A$$
∴ LHS = RHS. Hence proved. $\blacksquare$
(x) $\left(\dfrac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\dfrac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$
Part 1: We simplify $\dfrac{1 + \tan^2 A}{1 + \cot^2 A}$:
$$= \frac{\sec^2 A}{\cosec^2 A}$$$$= \frac{\sin^2 A}{\cos^2 A}$$
$$= \tan^2 A$$
Part 2: We simplify $\dfrac{1 - \tan A}{1 - \cot A}$:
$$= \frac{1 - \tan A}{1 - \frac{1}{\tan A}}$$$$= \frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}$$
$$= \frac{(1 - \tan A)\tan A}{-(1 - \tan A)}$$
$$= -\tan A$$
So:
$$\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = (-\tan A)^2 = \tan^2 A$$∴ Both expressions equal $\tan^2 A$. Hence proved. $\blacksquare$