NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2#

We find $\sin 60° \cos 30° + \sin 30° \cos 60°$:

$$= \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}$$

$$= \frac{3}{4} + \frac{1}{4}$$

$$= 1$$

∴ The value is 1.

We find $2\tan^2 45° + \cos^2 30° - \sin^2 60°$:

$$= 2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2$$

$$= 2 + \frac{3}{4} - \frac{3}{4}$$

$$= 2$$

∴ The value is 2.

We find $\dfrac{\cos 45°}{\sec 30° + \cosec 30°}$:

$$= \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$$

$$= \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}}$$

$$= \frac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})}$$

$$= \frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3} + 1)}$$

Multiplying the numerator and denominator by $(\sqrt{3} - 1)$:

$$= \frac{\sqrt{3}(\sqrt{3} - 1)}{2\sqrt{2}(3 - 1)}$$

$$= \frac{3 - \sqrt{3}}{4\sqrt{2}}$$

Multiplying the numerator and denominator by $\sqrt{2}$:

$$= \frac{3\sqrt{2} - \sqrt{6}}{8}$$

∴ The value is $\mathbf{\dfrac{3\sqrt{2} - \sqrt{6}}{8}}$.

We find $\dfrac{\sin 30° + \tan 45° - \cosec 60°}{\sec 30° + \cos 60° + \cot 45°}$:

$$= \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1}$$

$$= \frac{\frac{3}{2} - \frac{2}{\sqrt{3}}}{\frac{3}{2} + \frac{2}{\sqrt{3}}}$$

Multiplying the numerator and denominator by $2\sqrt{3}$:

$$= \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$$

Multiplying the numerator and denominator by $(3\sqrt{3} - 4)$:

$$= \frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - 4^2}$$

$$= \frac{27 - 24\sqrt{3} + 16}{27 - 16}$$

$$= \frac{43 - 24\sqrt{3}}{11}$$

∴ The value is $\mathbf{\dfrac{43 - 24\sqrt{3}}{11}}$.

We find $\dfrac{5\cos^2 60° + 4\sec^2 30° - \tan^2 45°}{\sin^2 30° + \cos^2 30°}$:

$$= \frac{5\left(\frac{1}{2}\right)^2 + 4\left(\frac{2}{\sqrt{3}}\right)^2 - (1)^2}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$$

$$= \frac{\frac{5}{4} + \frac{16}{3} - 1}{\frac{1}{4} + \frac{3}{4}}$$

$$= \frac{\frac{15 + 64 - 12}{12}}{1}$$

$$= \frac{67}{12}$$

∴ The value is $\mathbf{\dfrac{67}{12}}$.

We find $\dfrac{2\tan 30°}{1+\tan^2 30°}$:

$$= \frac{2 \times \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$$

$$= \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$$

$$= \frac{2}{\sqrt{3}} \times \frac{3}{4}$$

$$= \frac{\sqrt{3}}{2}$$$$\Rightarrow \frac{\sqrt{3}}{2} = \sin 60°$$

∴ Answer: (A) sin 60°

We find $\dfrac{1-\tan^2 45°}{1+\tan^2 45°}$:

$$= \frac{1 - 1}{1 + 1}$$

$$= 0$$

∴ Answer: (D) 0

Checking each option:

For A = 0°:

$$\sin 2A = \sin 0° = 0$$

$$2\sin A = 2 \sin 0° = 0$$

$$\Rightarrow 0 = 0$$

For A = 30°:

$$\sin 2A = \sin 60° = \frac{\sqrt{3}}{2}$$

$$2\sin A = 2 \sin 30° = 1$$

$$\Rightarrow \frac{\sqrt{3}}{2} \neq 1$$

For A = 45°:

$$\sin 2A = \sin 90° = 1$$

$$2\sin A = 2 \sin 45° = \sqrt{2}$$

$$\Rightarrow 1 \neq \sqrt{2}$$

For A = 60°:

$$2\sin A = 2 \sin 60° = \sqrt{3}$$

$$\Rightarrow \sqrt{3} > 1$$

The value of a sine is never more than 1, so sin 2A cannot equal $\sqrt{3}$.

∴ sin 2A = 2 sin A only when A = 0°. Answer: (A) 0°

We find $\dfrac{2\tan 30°}{1-\tan^2 30°}$:

$$= \frac{2 \times \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}}$$

$$= \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$$

$$= \frac{2}{\sqrt{3}} \times \frac{3}{2}$$

$$= \sqrt{3}$$$$\Rightarrow \sqrt{3} = \tan 60°$$

∴ Answer: (C) tan 60°

$\tan(A + B) = \sqrt{3}$ and $\tan 60° = \sqrt{3}$, so:

$$A + B = 60° \quad \cdots (1)$$

$\tan(A - B) = \dfrac{1}{\sqrt{3}}$ and $\tan 30° = \dfrac{1}{\sqrt{3}}$, so:

$$A - B = 30° \quad \cdots (2)$$

Adding equation (1) and equation (2):

$$\begin{array}{rcl} A + B &=& 60° \\ A - B &=& 30° \\ \hline 2A &=& 90° \end{array}$$

$$\Rightarrow A = 45°$$

Substituting $A = 45°$ in equation (1):

$$A + B = 60°$$

$$45° + B = 60°$$

$$\Rightarrow B = 15°$$

∴ $A = \mathbf{45°}$ and $B = \mathbf{15°}$

False.

Take A = 30° and B = 60°:

$$\sin(A + B) = \sin 90° = 1$$

$$\sin A + \sin B = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}$$$$\Rightarrow 1 \neq \frac{1 + \sqrt{3}}{2}$$

∴ sin (A + B) ≠ sin A + sin B in general.

True (for θ from 0° to 90°).

From the table:

$$\Rightarrow \sin 0° < \sin 30° < \sin 45° < \sin 60° < \sin 90°$$

That is, $0 < \dfrac{1}{2} < \dfrac{1}{\sqrt{2}} < \dfrac{\sqrt{3}}{2} < 1$.

∴ sin θ increases as θ increases.

False.

From the table:

$$\Rightarrow \cos 0° > \cos 30° > \cos 45° > \cos 60° > \cos 90°$$

That is, $1 > \dfrac{\sqrt{3}}{2} > \dfrac{1}{\sqrt{2}} > \dfrac{1}{2} > 0$.

∴ cos θ decreases as θ increases.

False.

Take θ = 30°:

$$\sin 30° = \frac{1}{2}$$

$$\cos 30° = \frac{\sqrt{3}}{2}$$$$\Rightarrow \frac{1}{2} \neq \frac{\sqrt{3}}{2}$$

∴ sin θ = cos θ is not true for all θ. It is true only for θ = 45°.

True.

$$\cot A = \frac{\cos A}{\sin A}$$

At A = 0°:

$$\cot 0° = \frac{\cos 0°}{\sin 0°} = \frac{1}{0}$$

Division by zero is not defined.

∴ cot A is not defined for A = 0°.