NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3#

sin A in terms of cot A:

Using $1 + \cot^2 A = \cosec^2 A$:

$$\cosec^2 A = 1 + \cot^2 A$$

$$\sin^2 A = \frac{1}{\cosec^2 A} = \frac{1}{1 + \cot^2 A}$$

$$\boxed{\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}}$$

sec A in terms of cot A:

$$\cos^2 A = 1 - \sin^2 A = 1 - \frac{1}{1+\cot^2 A} = \frac{\cot^2 A}{1 + \cot^2 A}$$

$$\sec^2 A = \frac{1+\cot^2 A}{\cot^2 A}$$

$$\boxed{\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}}$$

tan A in terms of cot A:

$$\boxed{\tan A = \frac{1}{\cot A}}$$

cos A:

$$\cos A = \frac{1}{\sec A}$$

sin A:

Using $\sin^2 A = 1 - \cos^2 A = 1 - \dfrac{1}{\sec^2 A} = \dfrac{\sec^2 A - 1}{\sec^2 A}$:

$$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$$

tan A:

$$\tan A = \frac{\sin A}{\cos A} = \frac{\sqrt{\sec^2 A - 1}}{\sec A} \times \sec A = \sqrt{\sec^2 A - 1}$$

cot A:

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

cosec A:

$$\cosec A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$$

$$9\sec^2 A - 9\tan^2 A = 9(\sec^2 A - \tan^2 A) = 9 \times 1 = \mathbf{9}$$

Answer: (B) 9

$$= \left(1 + \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta}\right)\left(1 + \frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta}\right)$$$$= \frac{(\cos\theta + \sin\theta + 1)}{\cos\theta} \cdot \frac{(\sin\theta + \cos\theta - 1)}{\sin\theta}$$$$= \frac{(\cos\theta + \sin\theta)^2 - 1}{\sin\theta\cos\theta}$$$$= \frac{\cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta - 1}{\sin\theta\cos\theta}$$$$= \frac{1 + 2\sin\theta\cos\theta - 1}{\sin\theta\cos\theta} = \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = \mathbf{2}$$

Answer: (C) 2

$$= \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 - \sin A)$$$$= \frac{(1 + \sin A)(1 - \sin A)}{\cos A}$$$$= \frac{1 - \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \mathbf{\cos A}$$

Answer: (D) cos A

$$= \frac{\sec^2 A}{\cosec^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$

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

$$\text{LHS} = \left(\frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta}\right)^2 = \frac{(1-\cos\theta)^2}{\sin^2\theta} = \frac{(1-\cos\theta)^2}{1-\cos^2\theta}$$$$= \frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)} = \frac{1-\cos\theta}{1+\cos\theta} = \text{RHS}$$

$$\text{LHS} = \frac{\cos^2 A + (1+\sin A)^2}{\cos A(1+\sin A)} = \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{\cos A(1+\sin A)}$$$$= \frac{1 + 1 + 2\sin A}{\cos A(1+\sin A)} = \frac{2(1+\sin A)}{\cos A(1+\sin A)} = \frac{2}{\cos A} = 2\sec A = \text{RHS}$$

Let $s = \sin\theta,\ c = \cos\theta$. Then $\tan\theta = s/c$ and $\cot\theta = c/s$.

$$\text{LHS} = \frac{s/c}{1 - c/s} + \frac{c/s}{1 - s/c} = \frac{s/c}{(s-c)/s} + \frac{c/s}{(c-s)/c}$$$$= \frac{s^2}{c(s-c)} + \frac{c^2}{s(c-s)} = \frac{s^2}{c(s-c)} - \frac{c^2}{s(s-c)} = \frac{s^3 - c^3}{sc(s-c)}$$$$= \frac{(s-c)(s^2+sc+c^2)}{sc(s-c)} = \frac{s^2+sc+c^2}{sc} = \frac{1+sc}{sc} = \frac{1}{sc} + 1 = 1 + \sec\theta\cosec\theta = \text{RHS}$$

$$\text{LHS} = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = 1 + \cos A$$$$\text{RHS} = \frac{\sin^2 A}{1-\cos A} = \frac{1-\cos^2 A}{1-\cos A} = \frac{(1-\cos A)(1+\cos A)}{1-\cos A} = 1 + \cos A$$

∴ LHS = RHS $\hspace{1cm}$ [Hence Proved]

Dividing numerator and denominator by $\sin A$:

$$\text{LHS} = \frac{\cot A - 1 + \cosec A}{\cot A + 1 - \cosec A}$$

Since $\cosec^2 A - \cot^2 A = 1$, i.e., $(\cosec A - \cot A)(\cosec A + \cot A) = 1$:

$$= \frac{(\cot A + \cosec A) - (\cosec^2 A - \cot^2 A)}{\cot A - \cosec A + 1}$$$$= \frac{(\cot A + \cosec A)[1 - (\cosec A - \cot A)]}{1 + \cot A - \cosec A}$$$$= \frac{(\cot A + \cosec A)(1 + \cot A - \cosec A)}{1 + \cot A - \cosec A} = \cosec A + \cot A = \text{RHS}$$

$$\text{LHS} = \sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}} = \sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}$$$$= \frac{1+\sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS}$$

$$\text{LHS} = \frac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(2\cos^2\theta - 1)}$$

Since $2\cos^2\theta - 1 = 2(1-\sin^2\theta) - 1 = 1 - 2\sin^2\theta$:

$$= \frac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(1 - 2\sin^2\theta)} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}$$

$$\text{LHS} = \sin^2 A + 2\sin A \cdot \cosec A + \cosec^2 A + \cos^2 A + 2\cos A \cdot \sec A + \sec^2 A$$$$= (\sin^2 A + \cos^2 A) + 2 + 2 + \cosec^2 A + \sec^2 A$$$$= 1 + 4 + (1 + \cot^2 A) + (1 + \tan^2 A)$$$$= 7 + \tan^2 A + \cot^2 A = \text{RHS}$$

$$\text{LHS} = \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) = \frac{1-\sin^2 A}{\sin A} \cdot \frac{1-\cos^2 A}{\cos A}$$$$= \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A} = \sin A \cos A$$$$\text{RHS} = \frac{1}{\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A}} = \frac{1}{\dfrac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \sin A \cos A$$

∴ LHS = RHS $\hspace{1cm}$ [Hence Proved]

Part 1: $\dfrac{1+\tan^2 A}{1+\cot^2 A} = \tan^2 A$

$$\frac{1+\tan^2 A}{1+\cot^2 A} = \frac{\sec^2 A}{\cosec^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \checkmark$$

Part 2: $\left(\dfrac{1-\tan A}{1-\cot A}\right)^2 = \tan^2 A$

$$\frac{1-\tan A}{1-\cot A} = \frac{1-\tan A}{1 - \dfrac{1}{\tan A}} = \frac{1-\tan A}{\dfrac{\tan A - 1}{\tan A}} = \frac{(1-\tan A)\tan A}{\tan A - 1} = \frac{-({\tan A - 1})\tan A}{\tan A - 1} = -\tan A$$$$\therefore \left(\frac{1-\tan A}{1-\cot A}\right)^2 = (-\tan A)^2 = \tan^2 A \checkmark$$