Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions (Exercise 3.2) with step-by-step explanations, formulas, and concept-based problem solving as per the latest CBSE and NCERT guidelines. These Class 11 Maths Exercise 3.2 solutions cover conversion of degree and radian measure, relation between radians and arc length, and help students build a strong foundation for higher-level mathematics and competitive exams like JEE, NDA, and CUET. All solutions are prepared by subject experts for easy understanding and accuracy. Click the print button to download study material and notes in PDF format.
Q.1 : Find the values of the other five trigonometric functions if $ \cos x = -\frac{1}{2} $, $x$ lies in third quadrant.
Solution :
$\cos x = -\frac{1}{2}$
$\sec x = \frac{1}{\cos x} = -2$
We know that
$$\cos^2 x + \sin^2 x = 1$$
$$\sin^2 x = 1 – \cos^2 x$$
$$\sin^2 x = 1 – \left(-\frac{1}{2}\right)^2$$
$$\sin^2 x = 1 – \frac{1}{4} = \frac{3}{4}$$
$$\sin x = \pm \frac{\sqrt{3}}{2}$$
Since $x$ lies in the third quadrant, $\sin x$ is negative.
$$\sin x = -\frac{\sqrt{3}}{2}$$
$$\csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$$
$$\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$$
$$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$$
Q.2 : Find the values of the other five trigonometric functions if $ \sin x = \frac{3}{5} $, $x$ lies in second quadrant.
Solution :
$\sin x = \frac{3}{5}$
$\csc x = \frac{1}{\sin x} = \frac{5}{3}$
We know that
$$\cos^2 x + \sin^2 x = 1$$
$$\cos^2 x = 1 – \sin^2 x$$
$$\cos^2 x = 1 – \left(\frac{3}{5}\right)^2 = 1 – \frac{9}{25} = \frac{16}{25}$$
$$\cos x = \pm \frac{4}{5}$$
Since $x$ lies in the second quadrant, $\cos x$ is negative.
$$\cos x = -\frac{4}{5}$$
$$\sec x = \frac{1}{\cos x} = -\frac{5}{4}$$
$$\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = -\frac{3}{4}$$
$$\cot x = \frac{1}{\tan x} = -\frac{4}{3}$$
Q.3 : Find the values of the other five trigonometric functions if $ \cot x = \frac{3}{4} $, $x$ lies in third quadrant.
Solution :
$\cot x = \frac{3}{4}$
$\tan x = \frac{1}{\cot x} = \frac{4}{3}$
We know that
$$1 + \tan^2 x = \sec^2 x$$
$$\sec^2 x = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$$
$$\sec x = \pm \frac{5}{3}$$
Since $x$ lies in the third quadrant, $\sec x$ is negative.
$$\sec x = -\frac{5}{3}$$
$$\cos x = \frac{1}{\sec x} = -\frac{3}{5}$$
We know that
$$1 + \cot^2 x = \csc^2 x$$
$$\csc^2 x = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{25}{16}$$
$$\csc x = \pm \frac{5}{4}$$
Since $x$ lies in the third quadrant, $\csc x$ is negative.
$$\csc x = -\frac{5}{4}$$
$$\sin x = \frac{1}{\csc x} = -\frac{4}{5}$$
Q.4 : Find the values of the other five trigonometric functions if $ \sec x = \frac{13}{5} $, $x$ lies in fourth quadrant.
Solution :
$\sec x = \frac{13}{5}$
$\cos x = \frac{1}{\sec x} = \frac{5}{13}$
We know that
$$1 + \tan^2 x = \sec^2 x$$
$$\tan^2 x = \sec^2 x – 1 = \left(\frac{13}{5}\right)^2 – 1 = \frac{169}{25} – 1 = \frac{144}{25}$$
$$\tan x = \pm \frac{12}{5}$$
Since $x$ lies in the fourth quadrant, $\tan x$ is negative.
$$\tan x = -\frac{12}{5}$$
$$\cot x = \frac{1}{\tan x} = -\frac{5}{12}$$
We know that
$$\sin^2 x + \cos^2 x = 1$$
$$\sin^2 x = 1 – \cos^2 x = 1 – \left(\frac{5}{13}\right)^2 = 1 – \frac{25}{169} = \frac{144}{169}$$
$$\sin x = \pm \frac{12}{13}$$
Since $x$ lies in the fourth quadrant, $\sin x$ is negative.
$$\sin x = -\frac{12}{13}$$
$$\csc x = \frac{1}{\sin x} = -\frac{13}{12}$$
Q.5 : Find the values of the other five trigonometric functions if $ \tan x = -\frac{5}{12} $, $x$ lies in second quadrant.
Solution :
$\tan x = -\frac{5}{12}$
$\cot x = \frac{1}{\tan x} = -\frac{12}{5}$
We know that
$$1 + \tan^2 x = \sec^2 x$$
$$\sec^2 x = 1 + \left(-\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144}$$
$$\sec x = \pm \frac{13}{12}$$
Since $x$ lies in the second quadrant, $\sec x$ is negative.
$$\sec x = -\frac{13}{12}$$
$$\cos x = \frac{1}{\sec x} = -\frac{12}{13}$$
We know that
$$1 + \cot^2 x = \csc^2 x$$
$$\csc^2 x = 1 + \left(-\frac{12}{5}\right)^2 = 1 + \frac{144}{25} = \frac{169}{25}$$
$$\csc x = \pm \frac{13}{5}$$
Since $x$ lies in the second quadrant, $\csc x$ is positive.
$$\csc x = \frac{13}{5}$$
$$\sin x = \frac{1}{\csc x} = \frac{5}{13}$$
Find the values of the Trigonometric Functions in Questions 6 to 10.
Q.6 : Find the value of $\sin 765^\circ$
Solution:
$\sin 765^\circ = \sin (720^\circ + 45^\circ)$
$= \sin (2 \times 360^\circ + 45^\circ)$
$= \sin (2\pi + 45^\circ)$
We know that $\sin x$ repeats its values after each $2\pi$ interval. Therefore,
$$\sin (2\pi + 45^\circ) = \sin 45^\circ$$
$$= \frac{1}{\sqrt{2}}$$
Hence, $\sin 765^\circ = \frac{1}{\sqrt{2}}.$
Q.7 : Find the value of $\csc (-1410^\circ)$
Solution:
$\csc (-1410^\circ) = \csc (-1440^\circ + 30^\circ)$
$= \csc (-4 \times 360^\circ + 30^\circ)$
$= \csc (-4\pi + 30^\circ)$
We know that $\csc x$ repeats its values after each $2\pi$ interval. Therefore,
$$\csc (-4\pi + 30^\circ) = \csc (2 \times 2\pi + (-4\pi + 30^\circ))$$
$$= \csc (30^\circ) = 2$$
Hence, $\csc (-1410^\circ) = 2.$
Q.8 : Find the value of $\tan \frac{19\pi}{3}$
Solution:
$\tan \frac{19\pi}{3} = \tan \left(\frac{18\pi}{3} + \frac{\pi}{3}\right)$
$= \tan (6\pi + \frac{\pi}{3})$
We know that $\tan x$ repeats its values after each $\pi$ interval. Therefore,
$$\tan (6\pi + \frac{\pi}{3}) = \tan \frac{\pi}{3} = \sqrt{3}$$
Hence, $\tan \frac{19\pi}{3} = \sqrt{3}.$
Q.9 : Find the value of $\sin \left(-\frac{11\pi}{3}\right)$
Solution:
$\sin \left(-\frac{11\pi}{3}\right) = \sin \left(-\frac{12\pi}{3} + \frac{\pi}{3}\right)$
$= \sin (-4\pi + \frac{\pi}{3})$
We know that $\sin x$ repeats its values after each $2\pi$ interval. Therefore,
$$\sin (-4\pi + \frac{\pi}{3}) = \sin (2 \times 2\pi + (-4\pi + \frac{\pi}{3}))$$
$$= \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$
Hence, $\sin \left(-\frac{11\pi}{3}\right) = \frac{\sqrt{3}}{2}.$
Q.7 : Find the value of $\cot \left(-\frac{15\pi}{4}\right)$
Solution:
$\cot \left(-\frac{15\pi}{4}\right) = \cot \left(-\frac{16\pi}{4} + \frac{\pi}{4}\right)$
$= \cot (-4\pi + \frac{\pi}{4})$
We know that $\cot x$ repeats its values after each $\pi$ interval. Therefore,
$$\cot (-4\pi + \frac{\pi}{4}) = \cot (4\pi + (-4\pi + \frac{\pi}{4}))$$
$$= \cot \frac{\pi}{4} = 1$$
Hence, $\cot \left(-\frac{15\pi}{4}\right) = 1.$
