NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

Question:2

Evaluate
$\cos\left [ \cos ^{-1} \left ( \frac{-\sqrt{3}}{2} \right ) +\frac{\pi}{6} \right]$

Answer:

We have
$\cos \left [ \cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right ) + \frac{\pi}{6}\right ]$
$\left [ Since,\cos\frac{5\pi}{6}= \frac{-\sqrt{3}}{2}\right ]$
$=\cos \left [ \cos^{-1} \left (\cos \frac{5\pi}{6} \right )+ \frac{\pi}{6}\right ]$
$=\cos\left ( \frac{5\pi}{6}+\frac{\pi}{6} \right )$

$\left [ since . \cos ^{-1}\left ( \cos x \right )=x;x\epsilon \left ( 0,\pi \right ) \right ]$
$= \cos \left ( \frac{6\pi}{6} \right )$
$= \cos \pi$
=-1

Question:3

Prove that $\cot \left ( \frac{\pi}{4}-2 \cot^{-1}3 \right )=7$

Answer:

We prove that
$\cot \left ( \frac{\pi}{4}-2 \cot^{-1}3 \right )=7$
$\Rightarrow \cot \left ( \frac{\pi}{4}-2 \cot^{-1}3 \right )=\cot^{-1}7$
$\Rightarrow 2\cot^{-1}3=\frac{\pi}{4}-\cot^{-1}7$
$\Rightarrow 2\tan^{-1}\frac{1}{3}=\frac{\pi}{4}-\tan^{-1}\frac{1}{7}$
$\Rightarrow 2\tan^{-1}\frac{1}{3}+\tan^{1}\frac{1}{7}=\frac{\pi}{4}$
$\left [ since , 2 \tan^{-1}(x)=2 tan^{-1}\frac{2x}{1-(x)^{2}} \right ]$
$\Rightarrow \tan^{-1}\frac{\frac{2}{3}}{1-\left ( \frac{1}{3} \right )^{2}}+\tan^{-1}\frac{1}{7}=\frac{\pi}{4}$
$\Rightarrow \tan^{-1}\left ( \frac{\frac{2}{3}}{\frac{8}{9}} \right )+\tan^{-1}\frac{1}{7}=\frac{\pi}{4}$
$\Rightarrow \tan^{-1}\left ( \frac{3}{4} \right )+\tan^{-1}\frac{1}{7}=\frac{\pi}{4}$
$\left [ since , \tan^{-1}x+ tan^{-1}y =\tan^{-1}\frac{x+y}{1-xy} \right ]$
$\Rightarrow \tan^{-1}\left ( \frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4},\frac{1}{7}} \right )=\frac{\pi}{4}$
$\Rightarrow \tan^{-1}\frac {\frac{\left (21+4 \right )}{28}}{\frac{\left ( 28-3 \right )}{28}}=\frac{\pi}{4}$
$\Rightarrow \tan^{-1}\frac {25}{25}=\frac{\pi}{4}$
$\Rightarrow \tan^{-1}\left ( 1 \right )=\frac{\pi}{4}$
$\Rightarrow 1=\tan\frac{\pi}{4}$
$\Rightarrow 1=1$
LHS=RHS
Hence Proved

Question:4

Find the value of $\tan ^{-1} \left ( -\frac{1}{\sqrt{3}} \right )+cot^{-1}\left ( \frac{1}{\sqrt{3}} \right )+tan^{-1}\left [ sin\left ( \frac{-\pi }{2} \right ) \right ]$

Answer:

We have
$\tan^{-1}\left ( -\frac{1}{\sqrt{3}} \right )+cot^{-1}\left ( \frac{1}{\sqrt{3}} \right )+ tan^{-1}\left [ sin\left ( \frac{-\pi }{2} \right ) \right ]$
$=\tan^{-1}\left (tan \frac{5\pi}{6} \right )+cot^{-1}\left (cot \frac{\pi}{3} \right )+ tan^{-1}\left (-1 \right )$
$=\tan^{-1}\left [ tan\left ( \pi-\frac{5\pi}{6} \right ) \right ]+\cot^{-1}\left ( \cot\frac{\pi}{3} \right )+\tan^{-1}\left [ \tan\left ( \pi-\frac{\pi}{4} \right ) \right ]$
$\left [ since, \tan^{-1}\left ( \tan x \right )=x, x\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right ); \cot^{-1}\left ( cot x \right )=x,x\epsilon \left ( 0,\pi \right ); and\: \tan^{-1}\left ( -x \right )=-\tan^{-1}x \right ]$
$=\tan^{-1}\left ( -\tan \frac{\pi}{6} \right )+\cot^{-1}\left ( \cot\frac{\pi}{3} \right )+\tan^{-1}\left [ -\tan\frac{\pi}{4} \right ]$
$=-\frac{\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{-2\pi+4\pi-3\pi}{12}$
$=-\frac{\pi}{12}$

Question:5

find the value of $\tan^{-1}\left ( tan\frac{2\pi}{3} \right )$

Answer:

We have
$\tan^{-1}\left ( \tan\frac{2\pi}{3} \right )=\tan^{-1}\tan\left ( \pi-\frac{\pi}{3} \right )$
$=\tan^{-1}\left ( -\tan\frac{\pi}{3} \right )$
$\left [ Since, \tan^{-1}\left ( -x \right )=-\tan^{-1}x,x\epsilon R \right ]$
$=-\tan^{-1}tan\frac{\pi}{3}$
$\left [ Since, \tan^{-1}\left ( \tan x \right ) =x,x\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )\right ]$
$=-\frac{\pi}{3}$

Question:6

show that $2\tan^{-1}\left ( -3 \right )=-\frac{\pi}{2}+\tan ^{-1}\left ( \frac{-4}{3} \right )$

Answer:

We have to prove ,
$2\tan^{-1}(-3)=-\frac{\pi}{2}+tan^{-1}\left ( \frac{-4}{3} \right )$
LHS=$2\tan^{-1}(-3)$ $\left [ Since, \tan^{-1}\left ( -x \right ) = -\tan^{-1}x,x\epsilon R\right ]$
$=-\left [ \cos^{-1}\frac{1-3^{2}}{1+3^{2}} \right ]$ $\left [ Since 2 \tan^{-1}x=\left [\cos^{-1}\frac{1-x^{2}}{1+x^{2}} \right ], x\geq 0\right ]$
$=-\left [ \cos^{-1}\left (\frac{-8}{10} \right ) \right ]$
$=-\left [ \cos^{-1}\left (\frac{-4}{5} \right )\right ]$
$=-\left [\pi- \cos^{-1}\left (\frac{4}{5} \right ) \right ]$ $=-\left [ since \cos^{-1}(-x)=\pi-\cos^{-1}x,x\epsilon \left [ -1,1 \right ] \right ]$
$=-\pi+\cos^{-1}\left ( \frac{4}{5} \right )$
$\left [ let \cos^{-1}\left ( \frac{4}{5} \right ) =0 \Rightarrow \cos \theta = \left ( \frac{4}{5} \right ) \Rightarrow \tan \theta = \left ( \frac{3}{4} \right ) \Rightarrow \theta = \tan^{-1} \left ( \frac{3}{4} \right )\right ]$
$=-\pi+\tan^{-1}\left (\frac{3}{4} \right )=-\pi+\left [ \frac{\pi}{2}-\cot^{-1}\left ( \frac{3}{4} \right ) \right ]$
$=-\frac{\pi}{2}-\cot^{-1}\left ( \frac{3}{4} \right )$
$\left [ Since, \tan^{-1}\left (-x \right )=-\tan^{-1}x \right ]$
$=-\frac{\pi}{2}+\tan^{-1}\left ( \frac{-4}{3} \right )$
$=-\frac{\pi}{2}+\tan^{-1}\left (- \frac{4}{3} \right )$
=RHS
Hence Proved.

Question:7

Find the real solution of the equation:
$\tan^{-1}\sqrt{x\left ( x+1 \right )}+\sin^{-1}\sqrt{x^{2}+x+1}=\frac{\pi}{2}$

Answer:

We have , $\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^{2}+x+1}=\frac{\pi}{2}.........(i)$
Let $\sin^{-1}\sqrt{x^{2}+x+1}=\theta$
$\Rightarrow \sin \theta =\sqrt{\frac{x^{2}+x+1}{1}}$
$\Rightarrow \tan \theta =\frac{\sqrt{x^{2}+x+1}}{\sqrt{-x^{2}-x}}$ $\left [ Since, \tan\theta =\frac{\sin \theta }{\cos \theta } \right ]$
$\Rightarrow \theta =\tan^{-1}\frac{\sqrt{x^{2}+x+1}}{\sqrt{-x^{2}-x}}=\sin^{-1}\sqrt{x^{2}+x+1}$
On Putting the value of $\theta$ in Eq. (i), We get
$\tan^{-1}\sqrt{x(x+1)}+\tan^{-1}\frac{\sqrt{x^{2}+x+1}}{\sqrt{-x^{2}-x}}=\frac{\pi}{2}.........(ii)$
we know that,
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy},xy< 1$
So,(ii) becomes,
$\tan^{-1}\left [ \frac{\sqrt{x\left ( x+1 \right )}+\frac{\sqrt{x^{2}+x+1}}{\sqrt{-x^{2}-x}}}{1-\sqrt{x\left ( x+1 \right )}\frac{\sqrt{x^{2}+x+1}}{\sqrt{-x^{2}-x}}} \right ]=\frac{\pi}{2}$
$\Rightarrow \tan^{-1}\left [ \frac{\sqrt{x\left ( x+1 \right )}+\frac{\sqrt{x^{2}+x+1}}{\sqrt{-1\left (x^{2}+x \right )}}}{1-\sqrt{x\left ( x+1 \right )}\frac{\sqrt{x^{2}+x+1}}{\sqrt{-1\left (x^{2}+x \right )}}} \right ]=\frac{\pi}{2}$
$\Rightarrow \frac{x^{2}+x+\sqrt{-\left ( x^{2}+x+1 \right )}}{\left [ 1-\sqrt{-\left ( x^{2}+x+1 \right ).\sqrt{\left ( x^{2}+x \right )}} \right ]}=\tan \frac{\pi}{2}=\frac{1}{0}$
$\Rightarrow \left [ 1-\sqrt{-\left ( x^{2}+x+1 \right )}.\sqrt{\left ( x^{2}+x \right )} \right ]=0$
$\Rightarrow -\left ( x^{2}+x+1 \right )=1\: or\: x^{2}+x=0$
$\Rightarrow x^{2}-x-1=1 \: or\: x\left ( x+1 \right )=0$
$\Rightarrow x^{2}+x+2=0 \: or\: x\left ( x+1 \right )=0$
$\Rightarrow x= \frac{-1\pm \sqrt{1-\left ( 4\times 2 \right )}}{2}\: or\: x=-1$
$\Rightarrow x= 0 \: or \: x=-1$
For real solution , we have x=0,-1.

Question:8

Find the value of $\sin\left ( 2 \tan^{-1}\frac{1}{3} \right )+\cos\left ( tan^{-1}2\sqrt{2} \right )$

Answer:

We have $\sin \left (2 \tan^{-1}\frac{1}{3} \right )+\cos \left ( tan^{-1}2\sqrt{2} \right )$
$=\sin \left [\sin^{-1}\left \{ \frac{2\times \frac{1}{3}}{1+\left ( \frac{1}{3} \right )^{2}} \right \} \right ]+\cos\left ( \cos^{-1}\frac{1}{3} \right )$
$\left [ Since, \: \tan^{-1}x =\cos^{-1}\frac{1}{\sqrt{1+x^{2}}}; 2\tan^{-1}\left ( x \right )=2 \tan^{-1}\frac{2x}{1-\left (x \right )^{2}}, -1\leq x\leq 1 \: and \: \tan^{-1}2\sqrt{2}=\cos^{-1}\frac{1}{3}\right ]$
=$\sin\left [ \sin^{-1}\left \{ \frac{\frac{2}{3}}{1+\frac{1}{9}} \right \} \right ]+\frac{1}{3}$ $\left [ Since, \cos\left ( \cos^{-1}x \right ) = x, x\epsilon \left \{ -1,1 \right \}\right ]$
$= \sin \left [ \sin^{-1}\left ( \frac{2\times 9}{3\times 10} \right ) \right ]+\frac{1}{3}$
$= \sin \left [ \sin^{-1}\left ( \frac{3}{5} \right ) \right ]+\frac{1}{3}$
$= \frac{3}{5}+\frac{1}{3}\left [ Since, \sin\left ( \sin^{-1}x \right )=x \right ]$
$= \frac{14}{15}$

Question:9

If $2 \tan ^{-1}\left ( \cos \theta \right )=\tan^{-1}\left ( cosec \theta \right ),$ then show that $\theta =\frac{\pi}{4}$, where n is any integer.

Answer:

We have $2 \tan^{-1}\left ( \cos \theta \right )=\tan^{-1}\left ( cosec\, \theta \right )$
$\Rightarrow \tan^{-1}\left ( \frac{2 \cos \theta }{1-\cos^{2}\theta } \right )=\tan^{-1}\left ( 2\, cosec\, \theta \right )$ $\left [ Since \: 2 \tan^{-1}x=\tan^{-1}\left ( \frac{2x}{1-x^{2}} \right ) \right ]$
$\Rightarrow \frac{2 \cos \theta }{\sin^{2}\theta }=2 \, cosec\, \theta$
$\Rightarrow \cot \theta . 2\, cosec\, \theta =2\, cosec\, \theta$
$\Rightarrow \cot \theta =1$
$\Rightarrow \cot \theta =\cot\frac{\pi}{4}$
$\Rightarrow \theta =\frac{\pi}{4}$
Hence Proved

Question:10

Show that $\cos \left ( 2 \tan^{-1}\frac{1}{7} \right )=\sin\left ( 4 \tan^{-1}\frac{1}{3} \right )$

Answer:

We have , $\cos\left ( 2 \tan^{-1}\frac{1}{7} \right )=\sin\left ( 4\tan^{-1}\frac{1}{3} \right )$
$\Rightarrow \cos\left [ \cos^{-1}\left ( \frac{1-\left ( \frac{1}{7} \right )^{2}}{1+\left ( \frac{1}{7} \right )^{2}} \right ) \right ]=sin\left ( 2.2\tan^{-1}\frac{1}{3} \right )$
$\left [ Since, 2 \tan^{-1}x=\left [ \cos^{-1}\frac{1-x^{2}}{1+x^{2}} \right ] ,x\geq 0 \right ]$
$\Rightarrow \cos\left [ \cos^{-1}\left ( \frac{\frac{48}{49}}{\frac{50}{49}} \right ) \right ]=\sin\left [ 2\left ( \tan^{-1}\frac{\frac{2}{3}}{1-\left ( \frac{1}{3} \right )^{2}} \right ) \right ]$
$\Rightarrow \cos \left [ \cos^{-1}\left ( \frac{24}{25} \right ) \right ]=\sin\left ( 2 \tan^{-1} \frac{3}{4} \right )$
$\Rightarrow \cos \left [ \cos^{-1}\left ( \frac{24}{25} \right ) \right ]=\sin\left ( \sin^{-1}\frac{2\times \frac{3}{4}}{1+\frac{9}{16}} \right )$ $\left [ Since , 2\tan^{-1}x=\tan^{-1}\left ( \frac{2x}{1-x^{2}} \right ) \right ]$
$\Rightarrow \frac{24}{25}=\sin\left ( \sin^{-1}\frac{\frac{3}{2}}{\frac{25}{16}} \right )$
$\Rightarrow \frac{24}{25}=\frac{48}{50}$
$\Rightarrow \frac{24}{25}=\frac{24}{25}$
Since LHS=RHS
Hence Proved

Question:11

Solve the following equation $\cos \left ( \tan^{-1}x \right )=\sin\left ( \cot^{-1}\frac{3}{4} \right )$

Answer:

We have $\cos \left ( \tan^{-1}x \right )=\sin\left ( \cot^{-1}\frac{3}{4} \right )$
$\Rightarrow \cos \left ( \cos^{-1}\frac{1}{\sqrt{x^{2}+2}} \right )=\sin\left ( \sin^{-1}\frac{4}{5} \right )........(i)$
Let $\tan^{-1}x=\theta _{1}\Rightarrow \tan\theta_{1}=\frac{x}{1}$
$\Rightarrow \cos \theta_{1}=\frac{1}{\sqrt{x^{2}+1}}.....(a)$
$\Rightarrow \theta_{1}=\cos^{-1}\frac{1}{\sqrt{x^{2}+1}}.....(c)$
And $\cot^{-1}=\theta_{2}\Rightarrow \cot^{-1}=\frac{3}{4}$
$\Rightarrow \sin \theta_{2}=\frac{4}{5}.......(b)$
$\Rightarrow \theta_{2}= \sin^{-1}\frac{4}{5}.......(d)$
From (c),(d);(i) becomes
$\Rightarrow \cos \theta_{1}= \sin\theta_{2}$
$\Rightarrow \frac{1}{\sqrt{x^{2}+1}}=\frac{4}{5}$ [From (a),(b)]
On squarinting both Sides, we get
$\Rightarrow 16\left (x^{2}+1 \right )=25$
$\Rightarrow 16x^{2}=9$
$\Rightarrow x^{2}=\left (\frac{3}{4} \right )^{2}$
$\Rightarrow x=\pm \frac{3}{4}$
$\therefore x=\frac{3}{4},-\frac{3}{4}.$

Question:12

Prove that $\tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^{2}$

Answer:

We have $\tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )=\frac{\pi}{4}+\frac{1}{2}cos^{-1}x^{2}$
LHS=$\tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )........(i)$
$\left [ x^{2}=\cos2\theta=\cos^{2}\theta+\sin^{2}\theta=1-2\sin^{2}\theta = 2\cos^{2}\theta-1 \right ]$
$\Rightarrow \cos^{-1}x^{2}=2\theta$
$\Rightarrow \theta=\frac{1}{2}\cos^{-1}x^{2}$
$\therefore \sqrt{1+x^{2}}=\sqrt{1+\cos2\theta}$
$\Rightarrow \sqrt{1+2\cos^{2}\theta-1}=\sqrt{2}\cos \theta$
And $\sqrt{1-x^{2}}=\sqrt{1-\cos2\theta}$
$\sqrt{1-1+2 \sin^{2}\theta}=\sqrt{2}\sin \theta$
$\therefore LHS = \tan^{-1}\left (\frac{ \sqrt{2}\cos \theta +\sqrt{2} \sin \theta}{ \sqrt{2}\cos \theta -\sqrt{2} \sin \theta}\right )$
$= \tan^{-1}\left (\frac{ \cos \theta + \sin \theta}{ \cos \theta - \sin \theta}\right )$
$= \tan^{-1}\left (\frac{1+\tan \theta}{ 1-\tan \theta}\right )$
$= \tan^{-1}\left \{\frac{\tan\left (\frac{\pi}{4} \right )+\tan \theta}{ \tan\left (\frac{\pi}{4} \right )-\tan \theta} \right \}$
$= \tan^{-1}\left [ \tan\left ( \frac{\pi}{4}+\theta \right ) \right ]$ $\left [ Since , \tan\left ( x+y \right )=\frac{\tan x+\tan y}{1-\tan x.\tan y} \right ]$
$=\frac{\pi}{4}+\theta$
$=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^{2}$
=RHS
LHS=RHS
Hence Proved

Question:13

Find the simplified from of $\cos^{-1}\left ( \frac{3}{5}\cos x +\frac{4}{5}\sin x \right )$ , where $x\epsilon \left [ \frac{-3\pi}{4},\frac{\pi}{4} \right ]$

Answer:

Let $\cos y=\frac{3}{5}$
$\Rightarrow \sin y=\frac{4}{5}$
$\Rightarrow y=\cos^{-1}\frac{3}{5}=\sin^{-1}\frac{4}{5}=\tan^{-1}\frac{4}{3}$
$\therefore \cos^{-1}\left [ \cos y. \cos x+\sin y. \sin x \right ]$
$\left [ since, \cos\left ( A-B \right ) = \cos A.\cos B + \sin A. \sin B \right ]$
$=\cos^{-1}\left [ \cos\left ( y-x \right ) \right ]$
$\left [ scine, \cos \left ( \cos^{-1}x \right )=x,x\epsilon \left \{ -1,1 \right \} \right ]$
=y-x
$\left [ scine, y=\tan^{-1}\frac{4}{3} \right ]$
$=\tan^{-1}\frac{4}{3} -x$

Question:14

Prove that $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}$

Answer:

we have $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}$
$LHS=\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}$

$let \: \: \sin^{-1}\frac{8}{17}=\theta_{1}$
$\Rightarrow \sin \theta_{1}=\frac{8}{17}$
$\Rightarrow \tan \theta_{1}=\frac{8}{15}\Rightarrow \theta_{1}=\tan^{-1}\frac{8}{15}$
And, $\sin \frac{3}{5}=\theta_{2}\Rightarrow \sin^{-1}\frac{3}{5}$
$\Rightarrow \tan \theta_{2}=\frac{3}{4}\Rightarrow \theta_{2}=\tan^{-1}\frac{3}{4}$
$=\tan^{-1}\frac{8}{15}+\tan^{-1}\frac{3}{4}$
$=\tan^{-1}\left [ \frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}} \right ]$ $\left [ Since , \tan^{-1}x+\tan^{-1} y=tan^{-1}\left ( \frac{x+y}{1-xy} \right ) \right ]$
$=\tan^{-1}\left [ \frac{\frac{77}{60}}{\frac{36}{60}} \right ]$
$=\tan^{-1}\left ( \frac{77}{36} \right )$
Let $=\theta _{3}=tan^{-1}\left ( \frac{77}{36} \right )\Rightarrow \tan \theta_{3}=\frac{77}{36}$
$\Rightarrow \sin \theta_{3}=\frac{77}{\sqrt{5929+1296}}=\frac{77}{85}$
$\therefore \theta _{3}=\sin^{-1}\left ( \frac{77}{85} \right )$
$= \sin^{-1}\left ( \frac{77}{85} \right )=RHS$
Hence proved

Question:15

Show that $\sin^{-1}\frac{5}{13}+\cos ^{-1}\frac{3}{5}=\tan^{-1}\frac{63}{16}$

Answer:

Solving LHS, $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
$Let \: \sin^{-1}\frac{5}{13}=x$
$\Rightarrow \sin x=\frac{5}{13}$
$And \, \cos^{2}x=1-\sin^{2}x$
$\Rightarrow 1-\frac{25}{169}=\frac{144}{169}$
$\Rightarrow \cos x= \sqrt{\frac{144}{169}}=\frac{12}{13}$
$\therefore \tan x=\frac{\sin x}{\cos x}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$
$\Rightarrow \tan x=\frac{5}{12}..........(i)$
Again , let $\cos^{-1}\frac{3}{5}=y$
$\Rightarrow \cos y=\frac{3}{5}$
$\therefore \sin y=\sqrt{1-\cos^{2}y}$
$\Rightarrow \sin y=\sqrt{1-\left (\frac{3}{5} \right )^{2}}$
$\Rightarrow \sin y=\sqrt{\frac{16}{25}}=\frac{4}{5}$
$\Rightarrow \tan y=\frac{\sin y}{\cos y}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}.........(ii)$
We know that, $\tan\left ( x+y \right )=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\frac{4}{3}}$ [from (i),(ii)]
$\Rightarrow \tan\left ( x+y \right )=\frac{\frac{15+48}{36}}{\frac{36-20}{36}}$
$\Rightarrow \tan\left ( x+y \right )=\frac{\frac{63}{36}}{\frac{16}{36}}$
$\Rightarrow \left ( x+y \right )=\tan^{-1}\frac{63}{16}$$\Rightarrow \left ( x+y \right )=\tan^{-1}\frac{63}{16}=RHS$
Since , LHS=RHS
Hence Proved.

Question:16

Prove that , $\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}=\sin^{-1}\frac{1}{\sqrt{5}}$

Answer:

Solving LHS, $\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}$
Let $\tan^{-1}\frac{1}{4}=x$
$\Rightarrow \tan x=\frac{1}{4}$
Squaring both sides,
$\Rightarrow \tan^{2} x=\frac{1}{16}$
$\Rightarrow \sec^{2} x-1=\frac{1}{16}$
$\Rightarrow \sec^{2} x=\frac{17}{16}$
$\Rightarrow \frac{1}{\cos^{2}x}=\frac{17}{16}$
$\Rightarrow \cos^{2}x=\frac{16}{17}$
$\Rightarrow \cos x=\frac{4}{\sqrt{17}}$
$Since,\: \sin^{2}x=1-\cos^{2}x$
$\Rightarrow \sin^{2}x=1-\frac{16}{17}=\frac{1}{17}$
$\Rightarrow \sin x=\frac{1}{\sqrt{17}}$
Again,
Let $\tan^{-1}\frac{2}{9}=y$
$\Rightarrow \tan y=\frac{2}{9}$
Squaring both sides,
$\Rightarrow \tan^{2}y=\frac{4}{81}$
$\Rightarrow \sec^{2}y-1=\frac{4}{81}$
$\Rightarrow \sec^{2}y=\frac{85}{81}$
$\Rightarrow \frac{1}{\cos^{2}y}=\frac{85}{81}$
$\Rightarrow \cos^{2}y=\frac{81}{85}$
$\Rightarrow \cos y=\frac{9}{\sqrt{85}}$
Since, $\sin^{2}y=1-\cos^{2}y$
$\Rightarrow \sin^{2}=1-\frac{81}{85}=\frac{4}{85}$
$\Rightarrow \sin x=\frac{2}{\sqrt{85}}$
We know that, $\sin(x+y)=\sin x.\sin y+ \cos x.\sin y$
$\Rightarrow \sin\left ( x+y \right )=\frac{1}{\sqrt{17}}.\frac{9}{\sqrt{85}}+ \frac{4}{\sqrt{17}}.\frac{2}{\sqrt{85}}$
$\Rightarrow \sin\left ( x+y \right )=\frac{17}{\sqrt{17}.\sqrt{85}}$
$\Rightarrow \sin\left ( x+y \right )=\frac{\sqrt{17}}{\sqrt{17}.\sqrt{5}}$
$\Rightarrow \sin\left ( x+y \right )=\frac{1}{\sqrt{5}}$
$\Rightarrow x+y =\sin^{-1}\frac{1}{\sqrt{5}}=RHS$
Since , LHS=RHS
Hence Proved

Question:17

Find the value of $4 \tan^{-1}\frac{1}{5}-tan^{-1}\frac{1}{239}$

Answer:

We have, $4 tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}$
$=2 \times \left ( 2 \tan^{-1}\frac{1}{5} \right )-\tan^{-1}\frac{1}{239}$
$=2 \left [ \tan^{-1}\frac{\frac{2}{5}}{1-\left ( \frac{1}{5} \right )^{2}} \right ]-\tan^{-1}\frac{1}{239}$ $\left [ since, 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-\left ( x \right )^{2}} \right ]$

$=2 \left [ \tan^{-1}\frac{\frac{2}{5}}{ \frac{24}{25} } \right ]-\tan^{-1}\frac{1}{239}$
$=2 \tan^{-1}\frac{5}{12}-\tan^{-1}\frac{1}{239}$
$=\left [ \tan^{-1}\frac{\frac{5}{6}}{1-\left (\frac{5}{12} \right )^{2}} \right ]-\tan^{-1}\frac{1}{239}$ $\left [ since, 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-\left ( x \right )^{2}} \right ]$
$=\tan^{-1}\frac{\frac{5}{6}}{1-\frac{25}{144}}-\tan^{-1}\frac{1}{139}$
$=\tan^{-1}\left ( \frac{144 \times 5}{119 \times 6} \right )-\tan^{-1}\frac{1}{239}$
$=\tan^{-1}\frac{120}{119}-\tan^{-1}\frac{1}{239}$
$=\tan^{-1}\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}.\frac{1}{239}}\left [ since, \tan^{-1} x-\tan^{-1}y=\tan^{-1}\left ( \frac{x-y}{1+xy} \right ) \right ]$
$=\tan^{-1}\left [\frac{28680-119}{28441+120} \right ]$
$=\tan^{-1}\frac{28561}{28561}$
$=\tan^{-1}\left ( 1 \right )$
$=\tan^{-1}\left ( \tan \frac{\pi}{4} \right )$
$=\frac{\pi}{4}$
Hence, $4 tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}=\frac{\pi}{4}$

Question:18

Show that $\tan \left ( \frac{1}{2}\sin^{-1}\frac{3}{4} \right )=\frac{4-\sqrt{7}}{3}$ and, justify why the other value $\frac{4+\sqrt{7}}{3}$ is ignored.

Answer:

Solving LHS,
$=\tan\left ( \frac{1}{2} \sin^{-1}\frac{3}{4} \right )$
$Let \frac{1}{2} \sin^{-1}\frac{3}{4} =\theta$
$\Rightarrow \sin^{-1}\frac{3}{4} =2\theta$
$\Rightarrow \frac{3}{4} =\sin2\theta$
$\Rightarrow \sin2\theta= \frac{3}{4}$
$\Rightarrow \frac{2 \tan \theta}{1+\tan^{2}\theta}= \frac{3}{4}$
$\Rightarrow 3+3 \tan^{2}\theta = 8 \tan \theta$
$\Rightarrow 3 \tan^{2}\theta - 8 \tan \theta =3$
$Let \tan \theta=y$
$\therefore 3y^{2}+8y+3=0$
$\Rightarrow y= \frac{8\pm \sqrt{64-4\times 3\times 3}}{2\times 3}$$\Rightarrow = \frac{8\pm \sqrt{28}}{6}$
$\Rightarrow y=\frac{2\left ( 4\pm \sqrt{7} \right )}{2\times 3}$
$\Rightarrow \tan \theta=\frac{\left ( 4\pm \sqrt{7} \right )}{3}$
$\Rightarrow \theta=\tan^{-1}\frac{\left ( 4\pm \sqrt{7} \right )}{3}$
{ but as we can see , $\frac{ 4+ \sqrt{7} }{3}> 1$, since $max\left [ \tan\left ( \frac{1}{2} \sin^{-1}\frac{3}{4}\right ) \right ]=1$}
$\tan\left ( \frac{1}{2} \sin^{-1}\frac{3}{4}\right ) =\frac{4-\sqrt{7}}{3}=RHS$
Note: Scince $-\frac{\pi}{2}\leq sin^{-1}\frac{3}{4}\leq \frac{\pi}{2}$
$\Rightarrow -\frac{\pi}{4}\leq \frac{1}{2}sin^{-1}\frac{3}{4}\leq \frac{\pi}{4}$
$\therefore \tan\left ( -\frac{\pi}{4} \right )\leq \tan\left ( \frac{1}{2}sin^{-1}\frac{3}{4} \right )\leq \tan\left ( \frac{\pi}{4} \right )$
$\Rightarrow -1\leq \tan\left ( \frac{1}{2}\sin^{-1}\frac{3}{4} \right )\leq 1$

Question:19

If a₁,a₂,a₃,.................a_n is an arithmetic progression with common difference d, then evaluate the following expression.
$\tan\left [ \tan^{-1}\left ( \frac{d}{1+a_{1}a_{2}} \right )+\tan^{-1}\left ( \frac{d}{1+a_{2}a_{3}} \right )+\tan^{-1}\left ( \frac{d}{1+a_{3}a_{4}} \right )+..........+\tan^{-1}\left ( \frac{d}{1+a_{n-1}a_{n}} \right ) \right ]$

Answer:

We have $a_{1}=a, a_{2}=a + d, a_{3}=a+2d.......$
And, $d=a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=......=a_{n}-a_{n-1}$
Given that,
$\tan\left [ \tan^{-1}\left (\frac{d}{1+a_{1}a_{2}}\right )+ \tan^{-1}\left (\frac{d}{1+a_{2}a_{3}}\right )+ \tan^{-1}\left (\frac{d}{1+a_{3}a_{4}}\right )+............+ \tan^{-1}\left (\frac{d}{1+a_{n-1}a_{n}}\right ) \right ]$
$=\tan^{-1}\left [ \tan^{-1}\left ( \frac{a_{2}-a_{1}}{1+a_{1}a_{2}} \right )+\tan^{-1}\left ( \frac{a_{3}-a_{2}}{1+a_{2}a_{3}} \right )+.........+\tan^{-1}\left ( \frac{a_{n}-a_{n-1}}{1+a_{n-1}a_{n}} \right ) \right ]$
$=\tan\left [ \left ( \tan^{-1}a_{2}-\tan^{-1}a_{1} \right ) +\left ( \tan^{-1}a_{3}-\tan^{-1}a_{2} \right ) +................+\left ( \tan^{-1}a_{n}-\tan^{-1}a_{n-1} \right ) \right ]$
$=\tan \left [ \tan^{-1}a_{n}-\tan^{-1}a_{1} \right ]$
$\left [ Scince , \tan^{-1}x-\tan^{-1}y=\tan^{-1}\left ( \frac{x-y}{1+xy} \right ) \right ]$
$=\tan\left [ \tan^{-1} \left ( \frac{a_{n}-a_{1}}{1+a_{1}a_{n}} \right )\right ]$
$\left [ scince, \tan\left ( \tan^{-1}x \right )=x \right ]$
$=\frac{a_{n}-a_{1}}{1+a_{1}a_{n}}$

Question:20

Which of the following in the principal value branch of $\cos^{-1}x$
A.$\left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]$
B.$\left ( 0,-\pi \right )$
C.$\left [ 0,\pi \right ]$
D.$\left ( 0,\pi \right )-\frac{\pi}{2}$

Answer:

Answer : (c)
We know that the principal value branch of $\cos^{-1}$ is $\left [ 0,\pi \right ]$

Question:21

Which of the following in the principal value branch of $cosec^{-1} x$ .
$A.\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
$B.\left [ 0,\pi \right ]-\left \{\frac{\pi}{2} \right \}$
$C.\left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]$
$D.\left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]-\left \{ 0 \right \}$

Answer:

Answer :(D)

We know that the principal value branch of $cosec^{-1}x$ is $\left [-\frac{\pi}{2} ,\frac{\pi}{2}\right ]-\left ( 0 \right )$

Question:22

If $3\tan^{-1}x+\cot^{-1}x=\pi$, then x equals to
A. 0
B. 1
C. -1
D. 1/2

Answer:

Answer : B
Given That, $3 \tan ^{-1}x+\cot^{-1}x=\pi$
$\Rightarrow 2 \tan^{-1}x+\tan^{-1}x+\cot^{-1}x=\pi$
$\Rightarrow 2 \tan^{-1}x=\pi-\frac{\pi}{2}$ $\left [ Scince, \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2} \right ]$
$\Rightarrow \tan^{-1}\frac{2x}{1-x^{2}}=\frac{\pi}{2}$ $\left [ Scince, 2tan^{-1}x=\tan^{-1}\frac{2x}{1-x^{2}} \right ]$
$\Rightarrow \frac{2x}{1-x^{2}}=\tan \frac{\pi}{2}$
$\Rightarrow \frac{2x}{1-x^{2}}=\tan \frac{1}{0}$
Cross multiplying
$\Rightarrow 1-x^{2}=0$
$\Rightarrow x^{2}=\pm 1$
Here only x=1 satifies the given equation.
Note:- Here ,Putting x=-1 in the given equation we get,
$3 \tan^{-1}(-1)+cot^{-1}(-1)=\pi$
$3 \tan^{-1} \left [ \tan\left (\frac{-\pi}{4} \right )\right ]+cot^{-1} \left [ \cot \left (\frac{-\pi}{4} \right )\right ]=\pi$
$3 \tan^{-1} \left [- \tan\left (\frac{\pi}{4} \right )\right ]+cot^{-1} \left [- \cot \left (\frac{\pi}{4} \right )\right ]=\pi$
$3 \tan^{-1} \left [\tan\left (\frac{\pi}{4} \right )\right ]+\pi-cot^{-1} \left [\cot \left (\frac{\pi}{4} \right )\right ]=\pi$
$-3\times \frac{\pi}{4}+\pi-\frac{\pi}{4}=\pi$
$-\pi+\pi=\pi$
$0\neq \pi$
Hence , x=-1 does not satisfy the given equation.

Question:23

The value of $\sin^{-1}\left [ cos\left ( \frac{33\pi}{5} \right ) \right ]$ is
$A. \frac{3\pi}{5}$
$B. \frac{-7\pi}{5}$
$C. \frac{\pi}{10}$
$D. \frac{-\pi}{10}$

Answer:

Answer :(D)
We have
$\sin^{-1}\left [ \cos\left ( \frac{33\pi}{5} \right ) \right ]$
$=\sin^{-1}\left [ \cos\left ( 6\pi+ \frac{3\pi}{5} \right ) \right ]$
$=\sin^{-1}\left [ \cos\left ( \frac{3\pi}{5} \right ) \right ]$ $\left [ Since , \cos\left ( 2n \pi+\theta \right ) = \cos\theta \right ]$
$=\sin^{-1}\left [ \cos \left ( \frac{\pi}{2}+\frac{\pi}{10} \right ) \right ]$
$=\sin^{-1}\left [ \sin \left (- \frac{\pi}{10} \right ) \right ]\left [ Since, \sin^{-1}\left ( x \right )=-\sin^{-1}x \right ]$
$=- \frac{\pi}{10} \left [ Since, \sin^{-1}\left ( \sin x \right )=-x, x\epsilon \left ( -\frac{\pi}{2} ,\frac{\pi}{2} \right ) \right ]$

Question:24

The domain of the function $\cos^{-1}\left ( 2x-1 \right )$ is
A.[0,1]
B.[-1,1]
C.(-1,1)
D.[0,$\pi$]

Answer:

Answer:(A)
We Have $f(x)=cos^{-1}\left ( 2x-1 \right )$
Scince $-1\leq 2x-1\leq 1$
$\Rightarrow 0\leq 2x\leq 2$
$\Rightarrow 0\leq x\leq 1$
$\therefore x\epsilon \left [ 0,1 \right ]$

Question:25

The domain of the function defined by $f(x)=\sin^{-1}\sqrt{x-1}$ is
A.[1,2]
B.[-1,1]
C.[0,1]
D. None of these

Answer:

Answer: (A)
$f(x)=\sin^{-1}\sqrt{x-1}$
$\Rightarrow 0\leq x-1\leq 1\left [ Since ,\sqrt{x-1}\geq 0 \, and\, -1\leq \sqrt{x-1} \leq 1\right ]$
$\Rightarrow 1\leq x\leq 2$
$\therefore x\epsilon \left [ 1,2 \right ]$

Question:26

If $\cos\left ( sin^{-1}\frac{2}{5} + cos^{-1}x \right )=0,$ then x is equal to
$A. \frac{1}{5}$
$B. \frac{2}{5}$
C.0
D.1

Answer:

Answer: (B)
Given, $\cos\left ( Sin^{-1}\frac{2}{5}+cos^{-1}x \right )=0$
Let $Sin^{-1}\frac{2}{5}+cos^{-1}x =\theta$
So $\cos \theta =0.......(1)$
Principal value $\cos^{-1} x$ is $\left [ 0,\pi \right ]$.......(2)
Also , we know that $\cos\frac{\pi}{2}=0......(3)$
From (1) ,,(2), and (3) we have
$\theta =\frac{\pi}{2}$
But $\theta =\sin^{-1}\frac{2}{5}+\cos^{-1}x$
So,
$\sin^{-1}\frac{2}{5}+\cos^{-1}x=\frac{\pi}{2}$
We know that $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} for\: all \: x\epsilon \left [ -1 ,1\right ]$
As $\sin^{-1}\frac{2}{5}+\cos^{-1}x=\frac{\pi}{2}$
so $x=\frac{2}{5}$

Question:27

The value of sin (2tan^–1 (.75)) is equal to
A. 0.75
B. 1.5
C. 0.96
D. sin 1.5

Answer:

Answer :(c)
sin (2tan^–1 (.75))
Let, tan^–1 (.75) = θ
$\Rightarrow \tan^{-1}\left (\frac{3}{4} \right )=\theta$
$\Rightarrow \tan \theta =\frac{3}{4}$
As, $\tan \theta =\frac{3}{4}$ so
$\sin \theta =\frac{3}{5}, \cos \theta =\frac{4}{5}......(1)$
Now,
sin (2tan^–1 (.75)) = sin 2θ
= 2 sin θ cos θ
$=2\left (\frac{3}{5} \right )\left (\frac{4}{5} \right )$
$=\frac{24}{25}$
So, sin (2tan^–1 (.75)) = 0.96.

Question:28

The Value of $\cos^{-1} \cos \frac{3\pi}{2}$ is equal to
$A. \frac{\pi}{2}$
$B. \frac{3\pi}{2}$
$C. \frac{5\pi}{2}$
$D. \frac{7\pi}{2}$

Answer:

We have $\cos^{-1}\cos\frac{3\pi}{2}$
We know that,
$\cos\frac{3\pi}{2}=0$
So, $\cos^{-1}\cos\frac{3\pi}{2}=\cos^{-1}0$
Let $\cos^{-1}0=\theta$
⇒ cos θ = 0
Principal value of cos^-1 x is [0, π]
For, cos θ = 0
so,$\theta=\frac{\pi}{2}$

Question:29

The value of the expression $2 \sec^{-1}2+\sin^{-1}\left ( \frac{1}{2} \right )$ is
$A.\frac{\pi}{6}$
$B.\frac{5\pi}{6}$
$C.\frac{7\pi}{6}$
D.1

Answer:

Answer :(B)
We have,
Principal value of sin^-1 x is $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
Principal value of sec^-1 x is [0, π]$-\left \{ \frac{\pi}{2} \right \}$
Let $\sin^{-1}\frac{1}{2}=A$
$\Rightarrow \sin A =\frac{1}{2}$
$\Rightarrow A =\frac{\pi}{6}$
So, $\Rightarrow \sin^{-1}\frac{1}{2}=\frac{\pi}{6}$ … (1)

Let sec^-1 2 = B
⇒ sec B = 2
$\Rightarrow B=\frac{\pi}{3}$
So, 2 sec^-1 2 = 2B
$\Rightarrow 2\sec^{-1}2=\frac{2\pi}{3}$...(2)
So, the value of $2\sec^{-1}2+\sin^{-1}\frac{1}{2}$ from (1) and (2) is
$2\sec^{-1}2+\sin^{-1}\frac{1}{2}=\frac{2\pi}{3}+\frac{\pi}{6}$
$=\frac{4\pi}{6}+\frac{\pi}{6}$
$=\frac{5\pi}{6}$
So, $2\sec^{-1}2+\sin^{-1}\frac{1}{2}=\frac{5\pi}{6}$

Question:30

If tan^–1 x + tan^–1 y = 4π/5, then cot^–1x + cot^–1 y equals
$A. \frac{\pi}{5}$
$B. \frac{2\pi}{5}$
$C. \frac{3\pi}{5}$
$D. \pi$

Answer:

Answer :(A)
We know that,
$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$
We have,
tan^–1 x + tan^–1 y = 4π/5 … (1)
Let, cot^–1x + cot^–1 y = k … (2)
Adding (1) and (2) –
$\tan^{-1}x+\tan^{-1}y+\cot^{-1}x+\cot^{-1}y=\frac{4\pi}{5}+k$...(3)
Now, tan^–1 A + cot^–1 A = π/2 for all real numbers.
So, (tan^–1 x + cot^–1 x) + (tan^–1y + cot^–1 y) = π … (4)
From (3) and (4), we get,
$\frac{4\pi}{5}+k=\pi$
$\Rightarrow k=\pi-\frac{4\pi}{5}$
$\Rightarrow k=\frac{\pi}{5}$

Question:31

If $\sin^{-1}\frac{2a}{1+a^{2}}+\cos ^{-1}\frac{1-a^{2}}{1+a^{2}}=tan^{-1}\frac{2x}{1-x^{2}}$ where a, $x\epsilon \left [ 0,1 \right ]$ then the value of x is
A. 0
B. a/2
C. a
D. $\frac{2a}{1-a^{2}}$

Answer:

Answer:(D)
We have
$sin^{-1}\frac{2a}{1+a^{2}}+cos^{-1}\frac{1-a^{2}}{1+a^{2}}=\tan^{-1}\frac{2x}{1-x^{2}}$
we know that
$2 \tan ^{-1}p=\sin^{-1}\frac{2p}{1+p^{2}}.........(1)$
$Also,2 \tan^{-1}p=\cos^{-1}\frac{1-p^{2}}{1+p^{2}}.........(2)$
$Also,2 \tan^{-1}p=\tan^{-1}\frac{2p}{1-p^{2}}.........(3)$
From (1) and (2) we have,
L.H.S-
$\sin^{-1}\frac{2a}{1+a^{2}}+\cos^{-1}\frac{1-a^{2}}{1+a^{2}}=2\tan^{-1}a+2\tan^{-1}a$
$\Rightarrow \sin^{-1}\frac{2a}{1+a^{2}}+\cos^{-1}\frac{1-a^{2}}{1+a^{2}}=4\tan^{-1}a$
From (3) R.H.S
$\tan^{-1}\frac{2x}{1-x^{2}}=2\tan^{-1}x$
So, we have 4 tan^-1 a = 2 tan^-1 x
⇒ 2 tan^-1 a = tan^-1 x
But from (3) $2\tan^{-1}a= \tan^{-1}\frac{2a}{1-a^{2}}$
So $\tan^{-1}\frac{2a}{1-a^{2}}=\tan^{-1}x$
$x=\frac{2a}{1-a^{2}}$

Question:32

The value of $\cot \cos^{-1}\frac{7}{25} is$
A. 25/24
B.25/7
C.24/25
D.7/24

Answer:

Answer :(d)
We have to find $\cot \cos^{-1}\frac{7}{25}$
Let $\cos^{-1}\frac{7}{25}=A$
$\Rightarrow \cos^{-1}=\frac{7}{25}$
Also, $\cot A=\cot \cos^{-1}\frac{7}{25}$
As, $\sin A=\sqrt{1-\cos^{2}A}$
So $\sin A=\sqrt{1-\left (\frac{7}{5} \right )^{2}}$
$\Rightarrow \sin A=\sqrt{1-\frac{49}{625} }$
$\Rightarrow \sin A=\sqrt{\frac{625-49}{625} }$
$\Rightarrow \sin A=\sqrt{\frac{576}{625} }$
$\Rightarrow \sin A={\frac{24}{25} }$
We need to find cot A
$\cot A=\frac{\cos A}{\sin A}$
$\Rightarrow \cot A=\frac{\frac{7}{25}}{\frac{24}{25}}$
$\Rightarrow \cot A=\frac{7}{24}$
So $\cot \cos^{-1}\frac{7}{25}=\frac{7}{24}$

Question:33

The value of the expression $\tan \frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}$ is $\left [ Hint: \tan\frac{\theta}{2} =\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right ]$
$A. 2+ \sqrt{5}$
$B.\sqrt{5}-2$
$C.\frac{2+\sqrt{5}}{2}$
$D. \sqrt{5}+2$
Answer:

Answer:(B)
We need to find , $\tan \frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}$
Let, $\cos^{-1}\frac{2}{\sqrt{5}}=A$
$\Rightarrow \cos A=\frac{2}{\sqrt{5}}$
Also we need to find $\tan\frac{A}{2}$
We know that $\tan\frac{\theta}{2}=\sqrt{\frac{\left ( 1-\cos \theta \right )}{1+\cos \theta}}$
so, $\tan^{-1}\frac{A}{2}=\sqrt{\frac{\left ( 1-\cos A \right )}{1+\cos A}}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{1-\frac{2}{\sqrt{5}}}{1+\frac{2}{\sqrt{5}}}}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{\frac{\sqrt{5}-2}{\sqrt{5}}}{\frac{\sqrt{5}+2}{\sqrt{5}}}}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}$
on rationalizing,
$\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{\left (\sqrt{5}-2 \right )\left ( \sqrt{5}+2 \right )}{\left (\sqrt{5}+2 \right )\left ( \sqrt{5}+2 \right )}}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{\left (\sqrt{5} \right )^{2}-2^{2}}{\left (\sqrt{5} ^{2}+2^{2}\right )}}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{5-4}{\left (\sqrt{5} ^{2}+2^{2}\right )}}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{1}{\left (\sqrt{5} ^{2}+2^{2}\right )}}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\frac{1}{\sqrt{5} +2}$
Again rationalizing
$\Rightarrow \tan^{-1} \frac{A}{2}=\frac{1\left ( \sqrt{5}-1 \right )}{\left (\sqrt{5} +2 \right )\left ( \sqrt{5}-2 \right )}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\frac{\left ( \sqrt{5}-2 \right)}{\left (\sqrt{5}^{2} -2^{2} \right )}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\frac{\left ( \sqrt{5}-2 \right)}{\left (5-4 \right )}$
$\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{5}-2$

Question:34

If |x| ≤ 1, then $2\tan ^{2}x+\sin^{-1}\frac{2x}{1+x^{2}}$ is equal to
A. 4 tan–1 x
B. 0
C. $\frac{\pi}{2}$
D. π

Answer:

Answer(A)
We need to find, $2 \tan^{-1}x+\sin^{-1}\frac{2x}{1+x^{2}}$
We know that
$2 \tan^{-1}p=\sin^{-1}\frac{2x}{1+x^{2}}$
So,
$2 \tan^{1}x+\sin^{-1}\frac{2x}{1+x^{2}}=2 \tan^{1}x+2\tan^{-1}x$
=$4 \tan^{1}x$

Question:35

If cos^–1α + cos^–1β + cos^–1γ = 3π, then α(β + γ) + β (γ + α) + γ (α + β) equals
A. 0
B. 1
C. 6
D. 12

Answer:

Answer :(C)
Given, cos^–1α + co^s–1β + cos^–1γ = 3π … (1)
Principal value of cos^-1 x is [0, π]
So, maximum value which cos^-1 x can have is π.
So, if (1) is correct then all the three terms i.e,
cos^–1α, cos^–1β, cos^–1γ should be equal to π
So, cos^–1α = π
cos^–1β = π
cos^–1γ = π
So, α = β = γ = -1
So, α(β + γ) + β (γ + α) + γ (α + β)
= (-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1)
= 3(-1)(-2)
= 6

Question:36

The number of real solutions of equarion $\sqrt{1+\cos x}=\sqrt{2}\cos^{-1}\left ( \cos x \right ) in \left [ \frac{\pi}{2},\pi \right ]$ is
A. 0
B. 1
C. 2
D. Infinite

Answer:

Answer:(A)
We have , $\sqrt{1+\cos 2x}=\sqrt{2}\cos^{-1}\left ( \cos x \right )$, x is in $\left [ \frac{\pi}{2}, \pi \right ]$
R.H.S
$\sqrt{2}\cos^{-1}\left ( \cos x \right )=\sqrt{2}x$
So, $\sqrt{1+\cos 2x}=\sqrt{2}x$
Squaring both side , we get,
$\left ( 1+\cos 2 x \right )=2x^{2}$
$\Rightarrow \cos 2x=2x^{2}-1$
Now plotting cos 2x and 2x²-1, we get,

As , there is no point of intersection in $\left [ \frac{\pi}{2},\pi \right ]$, so therre is no
solution of the given equation in $\left [ \frac{\pi}{2},\pi \right ]$

Question:37

If $\cos^{-1}x> \sin^{-1}x$ , then
$A. \frac{1}{\sqrt{2}}< x\leq 1$
$B. 0\leq x< \frac{1}{\sqrt{2}}$
$C.-1\leq x< \frac{1}{\sqrt{2}}$
D.x>0

Answer:

Answer :(C)
Plotting cos^-1 x and sin^-1 x, we get,

As, graph of cos^-1 x is above graph of sin^-1 x in $\left [ -1,\frac{1}{\sqrt{2}} \right )$.
So, cos^–1x > sin^–1 x for all x in $\left [ -1,\frac{1}{\sqrt{2}} \right )$ .

Question:38

Fill in the blanks The principle value of $\cos ^{-1}\left ( -\frac{1}{2} \right )$ is ___________.

Answer:

The principal value of $\cos^{-1}\left ( -\frac{1}{2} \right )$ is $\frac{2\pi}{3}$.
Principal value cos^-1 x is [0,$\pi$]
Let, $\cos^{-1}\left ( -1 \right )=\theta$
$\Rightarrow \cos \theta=-\frac{1}{2}$
As, $\cos \frac{2\pi}{3} =-\frac{1}{2}$
So, $\theta= \frac{2\pi}{3}$

Question:39

Fill in the blanks The value of $sin^{-1}\left ( sin \frac{3\pi}{5} \right )$ is_______.

Answer:

The value of $\sin^{-1}\left ( \sin\frac{3\pi}{5} \right )$ is $\frac{2\pi}{5}$
Principal value of $\sin^{-1}$ is $\left [ -\frac{\pi}{2} ,\frac{\pi}{2}\right ]$
now, $\sin^{-1}\left ( \sin\frac{3\pi}{5} \right )$ should be in the given range
$\frac{3\pi}{5}$ is outside the range $\left [ -\frac{\pi}{2} ,\frac{\pi}{2}\right ]$
As, sin (π – x) = sin x
So, $\sin^{-1}\left ( \sin\frac{3\pi}{5} \right )=\sin^{-1}\left ( \sin \left ( \pi-\frac{3\pi}{5} \right ) \right )$
$=\sin^{-1}\left ( \sin\frac{2\pi}{5} \right )$
$=\sin^{-1}\left ( \sin\frac{2\pi}{5} \right )=\frac{2\pi}{5}$

Question:40

Fill in the blanks
If cos (tan^–1x + cot^–1 √3) = 0, then value of x is _________.

Answer:

$\begin{aligned} &\text { If } \cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=0, \text { then value of } x \text { is } \sqrt{3} \text { . }\\ &\text { Given, } \cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=0\\ &\Rightarrow \tan ^{-1} x+\cot ^{-1} \sqrt{3}=\frac{\pi}{2}\\ &\text { We know that, } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\\ &\text { So, } x=\sqrt{3} \end{aligned}$

Question:40

Fill in the blanks
If cos (tan^–1x + cot^–1 √3) = 0, then value of x is _________.

Answer:

If cos (tan^–1x + cot^–1 √3) = 0, then value of x is $\sqrt{3}$
Given, cos (tan^–1x + cot^–1$\sqrt{3}$) = 0
$\Rightarrow \tan^{-1}x+\cot^{-1}\sqrt{3}=\frac{\pi}{2}$
we know that, $\Rightarrow \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$
so, x=$\sqrt{3}$

Question:41

fill in blanks the set of value of $\sec^{-1}\left (\frac{1}{2} \right )$ is______________.
Answer:

Fill in the blanks the set of value of $\sec^{-1}\left ( \frac{1}{2} \right )$ is $\phi$
Domain of sec^-1 x is R – (-1,1).
As, $-\frac{1}{2}$ is outside domain of sec^-1 x.
Which means there is no set of value of $\sec^{-1}\frac{1}{2}$
So, the solution set of $\sec^{-1}\frac{1}{2}$ is null set or $\phi$

Question:42

Fill in the blanks
The principal value of tan^–1 √3 is _________.

Answer:

The Principal value of $\tan^{-1} \sqrt{3}$ is $\frac{\pi}{3}$
Principal value of tan^-1 x is $\left (-\frac{\pi}{2},\frac{\pi}{2} \right )$
Let, $\tan^{-1}\left ( \sqrt{3} \right )=\theta$
$\Rightarrow \tan \theta=\sqrt{3}$
As $\Rightarrow \tan \frac{\pi}{3}=\sqrt{3}$
so, $\Rightarrow \theta=\sqrt{3}$

Question:43

The value of $\cos^{-1}\left ( \cos \frac{14\pi}{3} \right )$

Answer:

The value of $\cos^{-1}\left ( \cos \frac{14\pi}{3} \right )$ is $\frac{2\pi}{3}$
We needd, $\cos^{-1}\left ( \cos \frac{14\pi}{3} \right )$
Principal value of cos^-1 x is [0,π]
Also, cos (2nπ + θ) = cos θ for all n ? N
$\cos \frac{14\pi}{3}=\cos \left ( 4\pi+\frac{2\pi}{3} \right )$
$\Rightarrow \cos \frac{14\pi}{3}=\cos \frac{2\pi}{3}$
$So, \cos^{-1} \left (\cos \frac{14\pi}{3} \right )=\cos^{-1}\left (\cos \frac{2\pi}{3} \right )$
$\Rightarrow \cos^{-1} \left (\cos \frac{14\pi}{3} \right )=\frac{2\pi}{3}$

Question:44

Fill in the blanks
The value of cos (sin^–1 x + cos^–1 x), |x| ≤ 1 is ________.

Answer:

The value of cos (sin^–1 x + cos^–1 x) for |x| ≤ 1 is 0.
cos (sin^–1 x + cos^–1 x), |x| ≤ 1
We know that, (sin^–1 x + cos^–1 x), |x| ≤ 1 is $\frac{\pi}{2}$
So, $\cos\left ( \sin^{-1}x+\cos^{-1}x \right )=\cos\frac{\pi}{2}$
= 0

Question:45

The value of expression $\tan\left ( \frac{\sin^{-1}x+\cos^{-1}x }{2}\right ),$ when $x=\frac{\sqrt{3}}{2}$ is___________.

Answer:

The value of expression $\tan\left ( \frac{\sin^{-1}x+\cos^{-1}x}{2} \right )$ When $X=\frac{\sqrt{3}}{2}$ is 1
$\tan\left ( \frac{\sin^{-1}x+\cos^{-1}x}{2} \right )$ When $X=\frac{\sqrt{3}}{2}$
We know that, (sin^–1 x + cos^–1 x) for all |x| ≤ 1 is $\frac{\pi}{2}$
As, $x=\frac{\sqrt{3}}{2}$ lies in domain
So $\tan\left ( \frac{\sin^{-1}x+\cos^{-1}x}{2} \right )$=$\tan \frac{\pi}{4}$
=1

Question:46

Fill in the blanks if $y= 2 \tan^{-1}x+\sin^{-1}\frac{2x}{1+x^{2}}$ for all x, then ______<y<_____.

Answer:

Fill in the blanks if $y= 2 \tan^{-1}x+\sin^{-1}\frac{2x}{1+x^{2}}$ for all x, then $-2\pi< y< 2\pi$
$y= 2 \tan^{-1}x+\sin^{-1}\frac{2x}{1+x^{2}}$
We know that,
$2 \tan^{-1}p=\sin^{-1}\frac{2x}{1+x^{2}}$
so
$2 \tan^{1}x+\sin^{-1}\frac{2x}{1+x^{2}}=2 \tan^{1}x+2\tan^{-1}x$
=4 tan^-1 x
So, y = 4 tan^-1 x
As, principal value of tan-1 x is $\left (-\frac{\pi}{2},\frac{\pi}{2} \right )$
So, $4 \tan^{-1}x\epsilon \left ( -2\pi,2\pi \right )$
Hence, -2π < y < 2π

Question:47

The result $\tan^{-1}x-\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$ is true when value of xy is _________.

Answer:

The result $\tan^{-1}x-\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$ is true when value of xy is > -1.
We have,
$\tan^{-1}x-\tan^{-1}=\tan^{-1} \frac{x-y}{1+xy}$
Principal range of tan^-1a is $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
Let tan^-1x = A and tan^-1y = B … (1)
So, A,B $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
We know that, $\tan\left ( A-B \right )=\frac{\tan A - \tan B}{1-\tan A \tan B }$ … (2)
From (1) and (2), we get,
Applying, tan^-1 both sides, we get,
$\tan^{-1}\tan\left ( A-B \right )=\tan^{-1}\frac{x-y}{1-xy}$
As, principal range of tan^-1a is $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
So, for tan^-1tan(A-B) to be equal to A-B,
A-B must lie in $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$– (3)
Now, if both A,B < 0, then A, B $\epsilon \left ( -\frac{\pi}{2},0\right )$
∴ A $\epsilon \left ( -\frac{\pi}{2},0\right )$ and -B $\epsilon \left ( 0,\frac{\pi}{2}\right )$
So, A – B $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
So, from (3),
tan^-1tan(A-B) = A-B
$\Rightarrow \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{z+xy}$
Now, if both A,B > 0, then A, B $\epsilon \left ( 0,\frac{\pi}{2}\right )$
∴ A $\epsilon \left ( 0,\frac{\pi}{2}\right )$ and -B $\epsilon \left ( -\frac{\pi}{2},0\right )$
So, A – B $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
So, from (3),
tan^-1tan(A-B) = A-B
$\Rightarrow \tan{-1}x-\tan{-1}y=\tan^{-1}\frac{x-y}{z+xy}$
Now, if A > 0 and B < 0,
Then, A $\epsilon \left ( 0,\frac{\pi}{2}\right )$ and B $\epsilon \left ( 0,\frac{\pi}{2}\right )$
∴ A $\epsilon \left ( 0,\frac{\pi}{2}\right )$ and -B $\epsilon \left ( 0,\frac{\pi}{2}\right )$
So, A – B $\epsilon$ (0,π)
But, required condition is A – B $\epsilon$ $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
As, here A – B $\epsilon$ (0,π), so we must have A – B $\epsilon$ $\left ( 0,\frac{\pi}{2} \right )$
$A-B< \frac{\pi}{2}$
$A< \frac{\pi}{2} +B$
Applying tan on both sides,
$\tan A< \tan\left ( \frac{\pi}{2} +B \right )$
As, $\tan\left ( \frac{\pi}{2} +\alpha \right )=-\cot \alpha$
So, tan A < - cot B
Again, $\cot \alpha=\frac{1}{\tan \alpha}$
So, $\tan A< \frac{1}{\tan B}$
⇒ tan A tan B < -1
As, tan B < 0
xy > -1
Now, if A < 0 and B > 0,
Then, A $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$ and B $\epsilon$ $\left ( 0,\frac{\pi}{2} \right )$
∴ A $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$ and -B $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$
So, A – B $\epsilon$ (-π,0)
But, required condition is A – B $\epsilon$ $\left ( -\frac{\pi}{2} ,\frac{\pi}{2}\right )$
As, here A – B $\epsilon$ (0,π), so we must have A – B $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$
$\Rightarrow A-B> -\frac{\pi}{2}$
$\Rightarrow A>B -\frac{\pi}{2}$
Applying tan on both sides,
$\tan A>\tan\left (B -\frac{\pi}{2} \right )$
As, $\tan\left (\alpha -\frac{\pi}{2} \right )=-\cot \alpha$
So, tan B > - cot A
Again, $\cot \alpha\frac{1}{\tan \alpha}$
So, $\tan B >-\frac{1}{\tan A}$
⇒ tan A tan B > -1
⇒xy > -1

Question:48

Fill in the blanks
The value of cot^–1(–x) for all x ? R in terms of cot^–1 x is _______.

Answer:

The value of cot^–1(–x) for all x ? R in terms of cot^–1 x is
π – cot^-1 x.
Let cot^–1(–x) = A
⇒ cot A = -x
⇒ -cot A = x
⇒ cot (π – A) = x
⇒ (π – A) = cot^-1 x
⇒ A = π – cot-1 x
So, cot^–1(–x) = π – cot-1 x

Question:49

State True or False for the statement
All trigonometric functions have inverse over their respective domains.

Answer:

True.
It is well known that all trigonometric functions have inverse over their respective domains.

Question:50

State True or False for the statement
The value of the expression (cos^–1x)² is equal to sec² x.

Answer:

As, cos^-1 x is not equal to sec x. So, (cos^–1x)² is not equal to sec² x.

Question:51

State True or False for the statement
The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

Answer:

As, all trigonometric and their corresponding inverse functions are periodic so, we can obtain the inverse of a trigonometric ratio in any branch in which it is one-one and onto.

Question:52

State True or False for the statement
The least numerical value, either positive or negative of angle θ is called the principal value of the inverse trigonometric function.

Answer:

True
We know that the smallest value, either positive or negative of angle θ is called principal value of the inverse trigonometric function

Question:53

State True or False for the statement
The graph of an inverse trigonometric function can be obtained from the graph of their corresponding trigonometric function by interchanging x and y axes.

Answer:

True.
Graph of any inverse function can be obtained by interchanging x and y-axis in the graph of the corresponding function. If (p, q) are two points on f(x) then (q, p) will be on f^-1(x).

Question:54

State True or False for the statement
The minimum value of n for which $\tan^{-1}\frac{n}{\pi}>\frac{\pi}{4},n\epsilon N$ is valid is 5.

Answer:

false
$\tan ^{-1}\frac{n}{\pi}>\frac{\pi}{4}$
As , tan is an increasing function so applying tan on both side
we get,
$\tan\left (\tan ^{-1}\frac{n}{\pi} \right )>\tan \frac{\pi}{4}$
As, $\tan\left (\tan ^{-1}\frac{n}{\pi} \right )=\frac{n}{\pi}$ and $\tan\frac{\pi}{4}=1$
so $\frac{n}{\pi}>1$
⇒ n > π
⇒ n > 3.14
As, n is a natural number, so least value of n is 4.

Question:55

State True or False for the statement

The principal value of $\sin^{-1}\left [ \cos\left ( \sin^{-1}\frac{1}{2} \right ) \right ]$ is $\frac{\pi}{3}$

Answer:

True
Principal value of sin^-1 x is $\left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]$
Principal value of cos^-1 x is [0, π]
We have, $\sin^{-1}\left [ \cos \left [ \sin^{-1}\left (\frac{1}{2} \right ) \right ] \right ]$
As, $\sin\frac{\pi}{6}=\frac{1}{2}$ so
$\sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \cos\left [ \sin^{-1}\left (\sin \frac{\pi}{6} \right ) \right ] \right ]$
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \cos \left [ \frac{\pi}{6} \right ]\right ]$
As, $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ so,
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \sin \left [ \frac{\pi}{3} \right ]\right ]$
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\frac{\pi}{3}$