Class 11 Maths NCERT Solutions Chapter 3 Ex 3.2 – Trigonometric Functions of Any Angle

Given \cos x=-\dfrac{1}{2}.

\sec x=\dfrac{1}{\cos x}=-2

Using \sin^2x+\cos^2x=1:

\sin^2x=1-\cos^2x=1-\dfrac{1}{4}=\dfrac{3}{4}

\sin x=\pm\dfrac{\sqrt3}{2}

Since x lies in the third quadrant, sin x is negative (by the ASTC rule).

\sin x=-\dfrac{\sqrt3}{2}, which gives \text{cosec } x=-\dfrac{2}{\sqrt3}

Further:

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{-\sqrt3/2}{-1/2}=\sqrt3 and \cot x=\dfrac{\cos x}{\sin x}=\dfrac{1}{\sqrt3}

Given \sin x=\dfrac{3}{5}, so \text{cosec } x=\dfrac{5}{3}.

Using \sin^2x+\cos^2x=1:

\cos^2x=1-\dfrac{9}{25}=\dfrac{16}{25}

\cos x=\pm\dfrac{4}{5}

Since x lies in the second quadrant, cos x is negative.

\cos x=-\dfrac{4}{5}, which gives \sec x=-\dfrac{5}{4}

Further:

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{3/5}{-4/5}=-\dfrac{3}{4} and \cot x=-\dfrac{4}{3}

Given \cot x=\dfrac{3}{4}, so \tan x=\dfrac{4}{3}.

Using 1+\tan^2x=\sec^2x:

\sec^2x=1+\dfrac{16}{9}=\dfrac{25}{9}

\sec x=\pm\dfrac{5}{3}

Since x lies in the third quadrant, sec x is negative.

\sec x=-\dfrac{5}{3}, which gives \cos x=-\dfrac{3}{5}

Further, \sin x=\tan x\cos x=\dfrac{4}{3}\times\left(-\dfrac{3}{5}\right)=-\dfrac{4}{5}, which gives \text{cosec } x=-\dfrac{5}{4}

Given \sec x=\dfrac{13}{5}, so \cos x=\dfrac{5}{13}.

Using \sin^2x+\cos^2x=1:

\sin^2x=1-\dfrac{25}{169}=\dfrac{144}{169}

\sin x=\pm\dfrac{12}{13}

Since x lies in the fourth quadrant, sin x is negative.

\sin x=-\dfrac{12}{13}, which gives \text{cosec } x=-\dfrac{13}{12}

Further:

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{-12/13}{5/13}=-\dfrac{12}{5} and \cot x=-\dfrac{5}{12}

Given \tan x=-\dfrac{5}{12}.

Using 1+\tan^2x=\sec^2x:

\sec^2x=1+\dfrac{25}{144}=\dfrac{169}{144}

\sec x=\pm\dfrac{13}{12}

Since x lies in the second quadrant, sec x (like cos x) is negative.

\sec x=-\dfrac{13}{12}, which gives \cos x=-\dfrac{12}{13}

Further, \sin x=\tan x\cos x=\left(-\dfrac{5}{12}\right)\times\left(-\dfrac{12}{13}\right)=\dfrac{5}{13}, which gives \text{cosec } x=\dfrac{13}{5}

And \cot x=\dfrac{1}{\tan x}=-\dfrac{12}{5}

Since sin repeats after every 360°, subtract full revolutions:

765^\circ=2\times360^\circ+45^\circ

Therefore:

\sin765^\circ=\sin(2\times360^\circ+45^\circ)=\sin45^\circ

Add full revolutions of 360° until the angle lands between 0° and 360°:

-1410^\circ+4\times360^\circ=-1410^\circ+1440^\circ=30^\circ

Since cosec repeats after every 360°:

\text{cosec}(-1410^\circ)=\text{cosec}\,30^\circ

Since tan repeats after every \pi (not 2\pi), subtract multiples of \pi:

\dfrac{19\pi}{3}=6\pi+\dfrac{\pi}{3}   (since 6\pi=\dfrac{18\pi}{3}, and 6 is an integer multiple of \pi)

Therefore:

\tan\dfrac{19\pi}{3}=\tan\left(6\pi+\dfrac{\pi}{3}\right)=\tan\dfrac{\pi}{3}

Add a full revolution of 2\pi=\dfrac{6\pi}{3}, twice, until the angle lands within a familiar range:

-\dfrac{11\pi}{3}+2\times2\pi=-\dfrac{11\pi}{3}+\dfrac{12\pi}{3}=\dfrac{\pi}{3}

Since sin repeats after every 2\pi:

\sin\left(-\dfrac{11\pi}{3}\right)=\sin\dfrac{\pi}{3}

Since cot repeats after every \pi=\dfrac{4\pi}{4}, add multiples of \pi until the angle lands within a familiar range:

-\dfrac{15\pi}{4}+4\pi=-\dfrac{15\pi}{4}+\dfrac{16\pi}{4}=\dfrac{\pi}{4}

Therefore:

\cot\left(-\dfrac{15\pi}{4}\right)=\cot\dfrac{\pi}{4}