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 $\sin \left(2 \tan ^{-1}(.75)\right)$ is equal to
A. 0.75
B. 1.5
C. 0.96
D. sin 1.5

Answer:

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

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 $\cos^{-1}0=\theta$
$\Rightarrow \cos \theta=0$
Principal value of $\cos ^{-1} x$ is $[0, \pi]$
For, $\cos \theta=0$
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,
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \sin \left [ \frac{\pi}{3} \right ]\right ]$
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\frac{\pi}{3}$