Class 11 Maths Exercise 3.2: Trigonometric Signs and Values
Overview
This page provides comprehensive Class 11 Ch 3: Trigonometric Functions - Exercise 3.2 Practice. Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3-2. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
Trigonometric Functions: Signs and Values
About Class 11 Maths Trigonometric Functions Exercise 3.2
This NCERT Class 11 Maths exercise focuses on Trigonometric Functions: Signs and Values from the chapter Trigonometric Functions. Revise the relevant definitions, standard results, notation, and worked examples before attempting the questions.
Attempt each question independently and write the reasoning in complete steps. Check symbols, substitutions, diagrams, and final conditions carefully, then use the step-by-step solution to identify exactly what needs revision.
Q1
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Find the values of other five trigonometric functions if $\cos x = -\frac{1}{2}$, x lies in third quadrant.
Given $\cos x = -\frac{1}{2}$. Since x is in 3rd quadrant, $\cos x$ is negative (consistent).
$\sec x = \frac{1}{\cos x} = -2$.
$\sin^2 x + \cos^2 x = 1 \Rightarrow \sin^2 x = 1 - (-\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.
$\sin x = \pm \frac{\sqrt{3}}{2}$. Since x is in 3rd quadrant, $\sin x$ is negative. So $\sin x = -\frac{\sqrt{3}}{2}$.
$\csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$.
$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$.
$\sin x = -\frac{\sqrt{3}}{2}, \csc x = -\frac{2}{\sqrt{3}}, \sec x = -2, \tan x = \sqrt{3}, \cot x = \frac{1}{\sqrt{3}}$
Q2
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Find the values of other five trigonometric functions if $\sin x = \frac{3}{5}$, x lies in second quadrant.
Given $\sin x = \frac{3}{5}$.
$\csc x = \frac{1}{\sin x} = \frac{5}{3}$.
$\cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25}$.
$\cos x = \pm \frac{4}{5}$. Since x is in 2nd quadrant, $\cos x$ is negative. So $\cos x = -\frac{4}{5}$.
$\sec x = \frac{1}{\cos x} = -\frac{5}{4}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = -\frac{3}{4}$.
$\cot x = \frac{1}{\tan x} = -\frac{4}{3}$.
$\cos x = -\frac{4}{5}, \sec x = -\frac{5}{4}, \tan x = -\frac{3}{4}, \cot x = -\frac{4}{3}, \csc x = \frac{5}{3}$
Q3
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Find the values of other five trigonometric functions if $\cot x = \frac{3}{4}$, x lies in third quadrant.
Given $\cot x = \frac{3}{4}$.
$\tan x = \frac{1}{\cot x} = \frac{4}{3}$.
$\sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9}$.
$\sec x = \pm \frac{5}{3}$. Since x is in 3rd quadrant, $\sec x$ is negative. So $\sec x = -\frac{5}{3}$.
$\cos x = \frac{1}{\sec x} = -\frac{3}{5}$.
$\sin x = \tan x \cdot \cos x = \frac{4}{3} \cdot (-\frac{3}{5}) = -\frac{4}{5}$.
$\csc x = \frac{1}{\sin x} = -\frac{5}{4}$.
$\sin x = -\frac{4}{5}, \cos x = -\frac{3}{5}, \tan x = \frac{4}{3}, \sec x = -\frac{5}{3}, \csc x = -\frac{5}{4}$
Q4
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Find the values of other five trigonometric functions if $\sec x = \frac{13}{5}$, x lies in fourth quadrant.
Given $\sec x = \frac{13}{5}$.
$\cos x = \frac{1}{\sec x} = \frac{5}{13}$.
$\sin^2 x = 1 - \cos^2 x = 1 - \frac{25}{169} = \frac{144}{169}$.
$\sin x = \pm \frac{12}{13}$. Since x is in 4th quadrant, $\sin x$ is negative. So $\sin x = -\frac{12}{13}$.
$\csc x = \frac{1}{\sin x} = -\frac{13}{12}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = -\frac{12}{5}$.
$\cot x = \frac{1}{\tan x} = -\frac{5}{12}$.
$\sin x = -\frac{12}{13}, \cos x = \frac{5}{13}, \tan x = -\frac{12}{5}, \cot x = -\frac{5}{12}, \csc x = -\frac{13}{12}$
Q5
00:00
Find the values of other five trigonometric functions if $\tan x = -\frac{5}{12}$, x lies in second quadrant.
Given $\tan x = -\frac{5}{12}$.
$\cot x = -\frac{12}{5}$.
$\sec^2 x = 1 + \tan^2 x = 1 + \frac{25}{144} = \frac{169}{144}$.
$\sec x = \pm \frac{13}{12}$. Since x is in 2nd quadrant, $\sec x$ is negative. So $\sec x = -\frac{13}{12}$.
$\cos x = -\frac{12}{13}$.
$\sin x = \tan x \cdot \cos x = (-\frac{5}{12}) \cdot (-\frac{12}{13}) = \frac{5}{13}$.
$\csc x = \frac{13}{5}$.
$\sin x = \frac{5}{13}, \cos x = -\frac{12}{13}, \cot x = -\frac{12}{5}, \sec x = -\frac{13}{12}, \csc x = \frac{13}{5}$
Q6
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Find the value of the trigonometric function $\sin 765^\circ$.
We know that values of $\sin x$ repeat after an interval of $360^\circ$ or $2\pi$.
$765^\circ = 2 \times 360^\circ + 45^\circ$.
$\sin 765^\circ = \sin(2 \times 360^\circ + 45^\circ) = \sin 45^\circ$.
$\sin 45^\circ = \frac{1}{\sqrt{2}}$.
$\frac{1}{\sqrt{2}}$
Q7
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Find the value of the trigonometric function $\csc (-1410^\circ)$.
$\csc(-x) = -\csc x$. So $\csc(-1410^\circ) = -\csc(1410^\circ)$.
$1410^\circ = 4 \times 360^\circ - 30^\circ$ or $3 \times 360^\circ + 330^\circ$.
Let's use $1410^\circ = 4 \times 360^\circ - 30^\circ$.
$\csc(1410^\circ) = \csc(4 \times 360^\circ - 30^\circ) = \csc(-30^\circ) = -\csc 30^\circ$.
So, $-\csc(1410^\circ) = -(-\csc 30^\circ) = \csc 30^\circ = 2$.
Alternatively: $1410 = 3 \times 360 + 330$. $\csc(330) = \csc(360-30) = -\csc 30 = -2$. Then $-\csc(1410) = -(-2) = 2$.
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Q8
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Find the value of the trigonometric function $\tan \frac{19\pi}{3}$.
$\frac{19\pi}{3} = \frac{18\pi + \pi}{3} = 6\pi + \frac{\pi}{3}$.
Since $\tan x$ repeats after $\pi$, and $6\pi$ is a multiple of $2\pi$, the value is the same.
$\tan(6\pi + \frac{\pi}{3}) = \tan \frac{\pi}{3}$.
$\tan \frac{\pi}{3} = \sqrt{3}$.
$\sqrt{3}$
Q9
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Find the value of the trigonometric function $\sin (-\frac{11\pi}{3})$.
$\sin(-x) = -\sin x$. So $\sin(-\frac{11\pi}{3}) = -\sin(\frac{11\pi}{3})$.
$\frac{11\pi}{3} = \frac{12\pi - \pi}{3} = 4\pi - \frac{\pi}{3}$.
$\sin(4\pi - \frac{\pi}{3}) = \sin(-\frac{\pi}{3}) = -\sin \frac{\pi}{3}$.
So, $-\sin(\frac{11\pi}{3}) = -(-\sin \frac{\pi}{3}) = \sin \frac{\pi}{3}$.
$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.
$\frac{\sqrt{3}}{2}$
Q10
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Find the value of the trigonometric function $\cot (-\frac{15\pi}{4})$.
$\cot(-x) = -\cot x$. So $\cot(-\frac{15\pi}{4}) = -\cot(\frac{15\pi}{4})$.
$\frac{15\pi}{4} = \frac{16\pi - \pi}{4} = 4\pi - \frac{\pi}{4}$.
$\cot(4\pi - \frac{\pi}{4}) = \cot(-\frac{\pi}{4}) = -\cot \frac{\pi}{4}$.
So, $-\cot(\frac{15\pi}{4}) = -(-\cot \frac{\pi}{4}) = \cot \frac{\pi}{4}$.
$\cot \frac{\pi}{4} = 1$.
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