Last Updated on November 24, 2025 by Aman Singh
Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 3 Exercise 3.2. Learn to find the values of all five remaining trigonometric functions given one value and the quadrant of $x$. Master the use of trigonometric identities ($\sin^2 x + \cos^2 x = 1$) and the CAST rule to determine signs. Practice evaluating trig functions for large angles (e.g., $\sin 765^{\circ}$, $\tan(19\pi/3)$) using the periodicity of functions.
This exercise requires finding the values of the remaining five trigonometric functions given one function value and the quadrant of $x$, and evaluating trigonometric functions for large angles.
Finding the Values of Other Five Trigonometric Functions (Exercises 1-5)
We use the fundamental identities and the quadrant signs to determine the values.
1. $\cos x = -\frac{1}{2}$, $x$ lies in the third quadrant.
- $\sec x$ (Reciprocal of $\cos x$):$$\sec x = \frac{1}{\cos x} = \frac{1}{-1/2} = \mathbf{-2}$$
- $\sin x$ (Using Pythagorean Identity):$$\sin^2 x = 1 – \cos^2 x = 1 – \left(-\frac{1}{2}\right)^2 = 1 – \frac{1}{4} = \frac{3}{4}$$In the third quadrant, $\sin x$ is negative.$$\sin x = -\sqrt{\frac{3}{4}} = \mathbf{-\frac{\sqrt{3}}{2}}$$
- $\csc x$ (Reciprocal of $\sin x$):$$\csc x = \frac{1}{\sin x} = \frac{1}{-\sqrt{3}/2} = \mathbf{-\frac{2}{\sqrt{3}}}$$
- $\tan x$ (Quotient Identity):$$\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \mathbf{\sqrt{3}}$$
- $\cot x$ (Reciprocal of $\tan x$):$$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}} = \mathbf{\frac{1}{\sqrt{3}}}$$
| Function | Value |
| $\sin x$ | $\mathbf{-\sqrt{3}/2}$ |
| $\csc x$ | $\mathbf{-2/\sqrt{3}}$ |
| $\tan x$ | $\mathbf{\sqrt{3}}$ |
| $\cot x$ | $\mathbf{1/\sqrt{3}}$ |
| $\sec x$ | $\mathbf{-2}$ |
2. $\sin x = \frac{3}{5}$, $x$ lies in the second quadrant.
- $\csc x$ (Reciprocal of $\sin x$):$$\csc x = \frac{1}{\sin x} = \mathbf{\frac{5}{3}}$$
- $\cos x$ (Using Pythagorean Identity):$$\cos^2 x = 1 – \sin^2 x = 1 – \left(\frac{3}{5}\right)^2 = 1 – \frac{9}{25} = \frac{16}{25}$$In the second quadrant, $\cos x$ is negative.$$\cos x = -\sqrt{\frac{16}{25}} = \mathbf{-\frac{4}{5}}$$
- $\sec x$ (Reciprocal of $\cos x$):$$\sec x = \frac{1}{\cos x} = \mathbf{-\frac{5}{4}}$$
- $\tan x$ (Quotient Identity):$$\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = \mathbf{-\frac{3}{4}}$$
- $\cot x$ (Reciprocal of $\tan x$):$$\cot x = \frac{1}{\tan x} = \mathbf{-\frac{4}{3}}$$
3. $\cot x = \frac{4}{3}$, $x$ lies in the third quadrant.
- $\tan x$ (Reciprocal of $\cot x$):$$\tan x = \frac{1}{\cot x} = \mathbf{\frac{3}{4}}$$
- $\csc x$ (Using Pythagorean Identity: $1 + \cot^2 x = \csc^2 x$):$$\csc^2 x = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$$In the third quadrant, $\csc x$ is negative.$$\csc x = -\sqrt{\frac{25}{9}} = \mathbf{-\frac{5}{3}}$$
- $\sin x$ (Reciprocal of $\csc x$):$$\sin x = \frac{1}{\csc x} = \mathbf{-\frac{3}{5}}$$
- $\cos x$ (Using $\tan x$ and $\sin x$):Since $\tan x = \frac{\sin x}{\cos x}$, $\cos x = \frac{\sin x}{\tan x}$.$$\cos x = \frac{-3/5}{3/4} = -\frac{3}{5} \times \frac{4}{3} = \mathbf{-\frac{4}{5}}$$
- $\sec x$ (Reciprocal of $\cos x$):$$\sec x = \frac{1}{\cos x} = \mathbf{-\frac{5}{4}}$$
4. $\sec x = \frac{13}{5}$, $x$ lies in the fourth quadrant.
- $\cos x$ (Reciprocal of $\sec x$):$$\cos x = \frac{1}{\sec x} = \mathbf{\frac{5}{13}}$$
- $\sin x$ (Using Pythagorean Identity):$$\sin^2 x = 1 – \cos^2 x = 1 – \left(\frac{5}{13}\right)^2 = 1 – \frac{25}{169} = \frac{144}{169}$$In the fourth quadrant, $\sin x$ is negative.$$\sin x = -\sqrt{\frac{144}{169}} = \mathbf{-\frac{12}{13}}$$
- $\csc x$ (Reciprocal of $\sin x$):$$\csc x = \frac{1}{\sin x} = \mathbf{-\frac{13}{12}}$$
- $\tan x$ (Quotient Identity):$$\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = \mathbf{-\frac{12}{5}}$$
- $\cot x$ (Reciprocal of $\tan x$):$$\cot x = \frac{1}{\tan x} = \mathbf{-\frac{5}{12}}$$
5. $\tan x = -\frac{5}{12}$, $x$ lies in the second quadrant.
- $\cot x$ (Reciprocal of $\tan x$):$$\cot x = \frac{1}{\tan x} = \mathbf{-\frac{12}{5}}$$
- $\sec x$ (Using Pythagorean Identity: $1 + \tan^2 x = \sec^2 x$):$$\sec^2 x = 1 + \left(-\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144}$$In the second quadrant, $\sec x$ is negative.$$\sec x = -\sqrt{\frac{169}{144}} = \mathbf{-\frac{13}{12}}$$
- $\cos x$ (Reciprocal of $\sec x$):$$\cos x = \frac{1}{\sec x} = \mathbf{-\frac{12}{13}}$$
- $\sin x$ (Using $\tan x$ and $\cos x$):Since $\tan x = \frac{\sin x}{\cos x}$, $\sin x = \tan x \cdot \cos x$.$$\sin x = \left(-\frac{5}{12}\right) \cdot \left(-\frac{12}{13}\right) = \mathbf{\frac{5}{13}}$$
- $\csc x$ (Reciprocal of $\sin x$):$$\csc x = \frac{1}{\sin x} = \mathbf{\frac{13}{5}}$$
Finding the Values of Trigonometric Functions (Exercises 6-10)
We use the periodicity of trigonometric functions: $\sin(x + 2n\pi) = \sin x$, $\cos(x + 2n\pi) = \cos x$, and $\tan(x + n\pi) = \tan x$. For degrees, the period is $360^{\circ}$ (or $180^{\circ}$ for tangent/cotangent).
6. $\sin 765^{\circ}$
Divide $765^{\circ}$ by $360^{\circ}$: $765 = 2 \times 360 + 45$.
$$\sin 765^{\circ} = \sin (2 \times 360^{\circ} + 45^{\circ})$$
$$\sin 765^{\circ} = \sin 45^{\circ} = \mathbf{\frac{1}{\sqrt{2}}}$$
7. $\csc (-1410^{\circ})$
Use the odd function property: $\csc(-x) = -\csc x$.
$$\csc (-1410^{\circ}) = -\csc (1410^{\circ})$$
Divide $1410^{\circ}$ by $360^{\circ}$: $1410 = 3 \times 360 + 330$.
$$-\csc (1410^{\circ}) = -\csc (3 \times 360^{\circ} + 330^{\circ})$$
$$= -\csc (330^{\circ})$$
Use $\csc(360^{\circ} – \theta) = -\csc \theta$:
$$-\csc (360^{\circ} – 30^{\circ}) = -[-\csc 30^{\circ}]$$
$$= \csc 30^{\circ} = \mathbf{2}$$
8. $\tan \left(\frac{19\pi}{3}\right)$
Divide $19$ by $3$: $19/3 = 6 + 1/3$.
$$\tan \left(\frac{19\pi}{3}\right) = \tan \left(6\pi + \frac{\pi}{3}\right)$$
The period of $\tan x$ is $\pi$, so $\tan(6\pi + \theta) = \tan \theta$.
$$\tan \left(6\pi + \frac{\pi}{3}\right) = \tan \frac{\pi}{3} = \mathbf{\sqrt{3}}$$
9. $\sin \left(-\frac{11\pi}{3}\right)$
Use the odd function property: $\sin(-x) = -\sin x$.
$$\sin \left(-\frac{11\pi}{3}\right) = -\sin \left(\frac{11\pi}{3}\right)$$
Divide $11$ by $3$: $11/3 = 3 + 2/3$. We use $4\pi – \frac{\pi}{3}$.
$$-\sin \left(\frac{11\pi}{3}\right) = -\sin \left(4\pi – \frac{\pi}{3}\right)$$
Use $\sin(2n\pi – \theta) = -\sin \theta$.
$$-\sin \left(4\pi – \frac{\pi}{3}\right) = – \left[-\sin \left(\frac{\pi}{3}\right)\right]$$
$$= \sin \left(\frac{\pi}{3}\right) = \mathbf{\frac{\sqrt{3}}{2}}$$
10. $\cot \left(-\frac{15\pi}{4}\right)$
Use the odd function property: $\cot(-x) = -\cot x$.
$$\cot \left(-\frac{15\pi}{4}\right) = -\cot \left(\frac{15\pi}{4}\right)$$
Divide $15$ by $4$: $15/4 = 3 + 3/4$. We use $4\pi – \frac{\pi}{4}$.
$$-\cot \left(\frac{15\pi}{4}\right) = -\cot \left(4\pi – \frac{\pi}{4}\right)$$
The period of $\cot x$ is $\pi$, so $\cot(4\pi – \theta) = \cot(-\theta) = -\cot \theta$.
$$-\cot \left(4\pi – \frac{\pi}{4}\right) = – \left[-\cot \left(\frac{\pi}{4}\right)\right]$$
$$= \cot \left(\frac{\pi}{4}\right) = \mathbf{1}$$
