Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 8 Exercise 8.2.
Practice evaluating complex trigonometric expressions involving the standard angles $30^\circ, 45^\circ, 60^\circ$ (Q.1). Master algebraic manipulation and rationalization of results (Q.1, iv, v). Solutions include choosing and justifying the correct trigonometric option for simplified expressions (Q.2) and solving simultaneous equations to find angles A and B using inverse trigonometric values (Q.3). Critically assess and justify True/False statements regarding the behavior of $\sin \theta$ and $\cos \theta$ as the angle increases, and the definitions of trigonometric identities (Q.4). Essential practice for numerical application of specific trigonometric ratios.
This exercise requires knowledge of the trigonometric values for standard angles: $0^\circ, 30^\circ, 45^\circ, 60^\circ, \text{ and } 90^\circ$.
| θ | sinθ | cosθ | tanθ | cscθ | secθ | cotθ |
| $\mathbf{30^\circ}$ | $1/2$ | $\sqrt{3}/2$ | $1/\sqrt{3}$ | $2$ | $2/\sqrt{3}$ | $\sqrt{3}$ |
| $\mathbf{45^\circ}$ | $1/\sqrt{2}$ | $1/\sqrt{2}$ | $1$ | $\sqrt{2}$ | $\sqrt{2}$ | $1$ |
| $\mathbf{60^\circ}$ | $\sqrt{3}/2$ | $1/2$ | $\sqrt{3}$ | $2/\sqrt{3}$ | $2$ | $1/\sqrt{3}$ |
1. Evaluate the following
(i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
Using the standard values:
$$\left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = \mathbf{1}$$
(ii) $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$
Using the standard values:
$$2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 – \left(\frac{\sqrt{3}}{2}\right)^2 = 2(1) + \frac{3}{4} – \frac{3}{4} = 2 + 0 = \mathbf{2}$$
(iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}$
Substitute the values:
$$\frac{1/\sqrt{2}}{2/\sqrt{3} + 2} = \frac{1/\sqrt{2}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 + 2\sqrt{3}}$$
Rationalize the denominator:
$$\frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3} + 1)} \times \frac{\sqrt{3} – 1}{\sqrt{3} – 1}$$
$$= \frac{\sqrt{3}(\sqrt{3} – 1)}{2\sqrt{2}(3 – 1)} = \frac{3 – \sqrt{3}}{4\sqrt{2}}$$
Rationalize again to remove $\sqrt{2}$ from the denominator:
$$= \frac{3 – \sqrt{3}}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2} – \sqrt{6}}{8}$$
$$\mathbf{\frac{3\sqrt{2} – \sqrt{6}}{8}}$$
(iv) $\frac{\sin 30^\circ + \tan 45^\circ – \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
Substitute the values:
$$\frac{\frac{1}{2} + 1 – \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} = \frac{\frac{3}{2} – \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{3}{2}}$$
Find a common denominator ($2\sqrt{3}$) for the numerator and denominator:
$$= \frac{\frac{3\sqrt{3} – 4}{2\sqrt{3}}}{\frac{4 + 3\sqrt{3}}{2\sqrt{3}}} = \frac{3\sqrt{3} – 4}{3\sqrt{3} + 4}$$
Rationalize the denominator by multiplying by the conjugate $(3\sqrt{3} – 4)$:
$$= \frac{3\sqrt{3} – 4}{3\sqrt{3} + 4} \times \frac{3\sqrt{3} – 4}{3\sqrt{3} – 4} = \frac{(3\sqrt{3} – 4)^2}{(3\sqrt{3})^2 – 4^2}$$
$$= \frac{(9 \times 3) – 2(3\sqrt{3})(4) + 16}{27 – 16} = \frac{27 – 24\sqrt{3} + 16}{11} = \mathbf{\frac{43 – 24\sqrt{3}}{11}}$$
(v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
Note: The denominator simplifies to $1$, since $\sin^2 \theta + \cos^2 \theta = 1$.
$$\text{Denominator} = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$$
Now evaluate the numerator:
$$\text{Numerator} = 5\left(\frac{1}{2}\right)^2 + 4\left(\frac{2}{\sqrt{3}}\right)^2 – (1)^2$$
$$\text{Numerator} = 5\left(\frac{1}{4}\right) + 4\left(\frac{4}{3}\right) – 1 = \frac{5}{4} + \frac{16}{3} – 1$$
Find a common denominator (12):
$$\text{Numerator} = \frac{5(3)}{12} + \frac{16(4)}{12} – \frac{12}{12} = \frac{15 + 64 – 12}{12} = \frac{67}{12}$$
$$\text{Value} = \frac{67/12}{1} = \mathbf{\frac{67}{12}}$$
2. Choose the correct option and justify your choice
(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$
$$\frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3}$$
$$= \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{3\sqrt{3}}{2(3)} = \frac{\sqrt{3}}{2}$$
Since $\sin 60^\circ = \sqrt{3}/2$, the correct option is (A) $\sin 60^\circ$.
(ii) $\frac{1 – \tan^2 45^\circ}{1 + \tan^2 45^\circ}$
$$\frac{1 – (1)^2}{1 + (1)^2} = \frac{1 – 1}{1 + 1} = \frac{0}{2} = 0$$
Since $\sin 90^\circ = 1$, $\cos 90^\circ = 0$. $\sin 45^\circ = 1/\sqrt{2}$. $\tan 90^\circ$ is undefined.
The value is 0. The correct option is (D) $0$.
(iii) $\sin 2A = 2 \sin A$ is true when $A = $
We test the options:
- (A) $A = 0^\circ$: $\sin(2 \cdot 0^\circ) = \sin 0^\circ = 0$. $2 \sin 0^\circ = 2(0) = 0$. (True)
- (B) $A = 30^\circ$: $\sin(2 \cdot 30^\circ) = \sin 60^\circ = \sqrt{3}/2$. $2 \sin 30^\circ = 2(1/2) = 1$. ($\sqrt{3}/2 \neq 1$)The identity is only true when $A=0^\circ$.The correct option is (A) $0^\circ$.
(iv) $\frac{2 \tan 30^\circ}{1 – \tan^2 30^\circ}$
$$\frac{2(1/\sqrt{3})}{1 – (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 – 1/3} = \frac{2/\sqrt{3}}{2/3}$$
$$= \frac{2}{\sqrt{3}} \times \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3}$$
Since $\tan 60^\circ = \sqrt{3}$, the correct option is (C) $\tan 60^\circ$.
3. Find $A$ and $B$
Given:
- $\tan (A + B) = \sqrt{3}$
- $\tan (A – B) = 1/\sqrt{3}$
We know that $\tan 60^\circ = \sqrt{3}$ and $\tan 30^\circ = 1/\sqrt{3}$.
$$\tan (A + B) = \tan 60^\circ \implies A + B = 60^\circ \quad \dots (1)$$
$$\tan (A – B) = \tan 30^\circ \implies A – B = 30^\circ \quad \dots (2)$$
Add (1) and (2):
$$(A + B) + (A – B) = 60^\circ + 30^\circ$$
$$2A = 90^\circ \implies A = 45^\circ$$
Substitute $A = 45^\circ$ into (1):
$$45^\circ + B = 60^\circ \implies B = 60^\circ – 45^\circ = 15^\circ$$
Check constraints: $0^\circ < 45^\circ + 15^\circ = 60^\circ \le 90^\circ$ (True), and $45^\circ > 15^\circ$ (True).
Thus, $\mathbf{A = 45^\circ}$ and $\mathbf{B = 15^\circ}$.
4. State whether the following are true or false. Justify your answer.
(i) $\sin (A + B) = \sin A + \sin B$.
False.
Justification: Let $A = 60^\circ$ and $B = 30^\circ$.
- $\text{LHS} = \sin(60^\circ + 30^\circ) = \sin 90^\circ = 1$.
- $\text{RHS} = \sin 60^\circ + \sin 30^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2}$.Since $1 \neq (\sqrt{3} + 1)/2$, the statement is false.
(ii) The value of $\sin \theta$ increases as $\theta$ increases.
True.
Justification: By observing the standard values or the unit circle, as $\theta$ increases from $0^\circ$ to $90^\circ$:
- $\sin 0^\circ = 0$
- $\sin 30^\circ = 0.5$
- $\sin 45^\circ \approx 0.707$
- $\sin 60^\circ \approx 0.866$
- $\sin 90^\circ = 1$The value of $\sin \theta$ monotonically increases in the interval $0^\circ \le \theta \le 90^\circ$.
(iii) The value of $\cos \theta$ increases as $\theta$ increases.
False.
Justification: As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\cos \theta$ decreases:
- $\cos 0^\circ = 1$
- $\cos 30^\circ \approx 0.866$
- $\cos 60^\circ = 0.5$
- $\cos 90^\circ = 0$The value of $\cos \theta$ monotonically decreases in the interval $0^\circ \le \theta \le 90^\circ$.
(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
False.
Justification: This is only true when $\theta = 45^\circ$. For example, if $\theta = 30^\circ$, $\sin 30^\circ = 1/2$ and $\cos 30^\circ = \sqrt{3}/2$. Since $1/2 \neq \sqrt{3}/2$, the statement is false.
(v) $\cot A$ is not defined for $A = 0^\circ$.
True.
Justification: $\cot A = \frac{\cos A}{\sin A}$.
For $A = 0^\circ$, $\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0}$. Division by zero is undefined.
Last Updated on November 28, 2025 by Aman Singh
