NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

Edited By Komal Miglani | Updated on Mar 31, 2025 03:20 AM IST | #CBSE Class 12th

Imagine trying to find an unknown angle when only its sine or cosine value is given. Have you ever wondered how calculators determine the angles for trigonometric values? This is where inverse trigonometric functions come into play. They help us reverse the process of basic trigonometric functions like sine, cosine, and tangent, allowing us to find angles from given ratios. When the trigonometric values are known, the inverse trigonometric functions allow you to find the angles. These functions are very important when it comes to calculus, solving equations, and even putting them into practice in the real world, such as physics and engineering. We will go through all the NCERT Exemplar questions and solutions in this chapter, which are aimed at developing a solid understanding of this topic to help students understand what the author is trying to assess in the exam.

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  2. Sub-Topics Covered in NCERT Exemplar Class 12 Maths Solutions Chapter 2
  3. NCERT Exemplar Class 12 Maths Solutions
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  5. NCERT solutions of class 12 - Subject-wise
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  7. NCERT Books and NCERT Syllabus
  8. NCERT Exemplar Class 12 Solutions - Subject Wise

In NCERT Solutions for Class 12 Maths Chapter 2, students will understand how to find the ranges and domains of inverse trigonometric functions. Some other key concepts that students will learn are the behavior of the function, properties of inverse trigonometric functions, and more, all to be explained in a systematic way. The NCERT Exemplar Class 12 Maths Chapter 2 Solutions will provide step-by-step guidance to ensure a strong conceptual understanding for students.

NCERT Exemplar Class 12 Maths Solutions Chapter 2

Class 12 Maths Chapter 2 exemplar solutions Exercise: 2.3
Page number: 35-41
Total questions: 55
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Question:1

find the value of $\tan ^{-1}\left ( \tan \frac{5\pi }{6} \right )+\cos ^{-1}\left ( \cos \frac{13\pi }{6} \right )$

Answer:

we know that
$\tan^{-1}\left ( \tan x \right )=x; x \epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$ and
$\cos^{-1}\left ( \cos x \right )=x; x \epsilon \left ( 0,\pi \right)$

$\therefore \tan^{-1} \left ( \tan \frac{5\pi}{6} \right )+\cos^{-1}\left ( \cos \frac{13\pi}{6} \right )$
$= \tan^{-1}\left [ \tan \left ( \pi-\frac{\pi}{6} \right ) \right ]+\cos^{-1}\left [ cos\left ( \pi +\frac{7\pi}{6} \right ) \right ]$
$= \tan^{-1}\left (-\tan \frac{\pi}{6} \right )+cos^{-1}\left (-\cos\frac{7\pi}{6} \right )$ [since , $\cos\left ( \pi +\theta \right )=-\cos \theta$ ]
$=- \tan^{-1}\left (\tan \frac{\pi}{6} \right )+ \pi-\left [cos^{-1}\left (\cos\frac{7\pi}{6} \right ) \right ]$
$\left [ since \, \tan^{-1}\left ( -x \right )=- \tan 1 x , x\epsilon R \, and \, \cos^{-1}=\left ( -x \right )=\pi-\cos^{-1}\left ( x \right ), x\epsilon \left ( -1,1 \right ) \right ]$
$=- \tan^{-1}\left (\tan \frac{\pi}{6} \right )+ \pi-\cos^{-1}\left [\cos\left (\pi+\frac{\pi}{6} \right )\right ]$
$=- \tan^{-1}\left (\tan \frac{\pi}{6} \right )+ \pi - \cos ^{-1} \left (-\cos\frac{\pi}{6} \right )$ [since , $\cos\left ( \pi +\theta \right )=-\cos \theta$ ]
$=- \tan^{-1}\left (\tan \frac{\pi}{6} \right )+ \pi - \pi +\cos ^{-1} \left (\cos\frac{\pi}{6} \right )$
$=-\frac{\pi}{6}+0+\frac{\pi}{6}$
$=0$

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,.................an 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 + cos–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,
a36
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,
a37
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,
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \sin \left [ \frac{\pi}{3} \right ]\right ]$
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\frac{\pi}{3}$

Sub-Topics Covered in NCERT Exemplar Class 12 Maths Solutions Chapter 2

The sub-topics that are covered in this chapter of inverse trigonometric functions are:

  • Introduction
  • Basic concepts
  • Properties of inverse trigonometric functions
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NCERT Exemplar Class 12 Maths Solutions

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 2

Class 12 Math NCERT Exemplar solutions chapter 2, students will get detailed answers to the questions in the NCERT book after every topic. Understanding and grasping this chapter can help one aim for a better score in their school exams, boards and their entrance exams.

  • In NCERT Exemplar solutions for Class 12 Math chapter 2, cover properties and graphical representations of inverse trigonometric functions.
  • One will learn about the necessity of studying inverse trigonometric functions and their properties. It covers the basic details about inverse trigonometric functions.
  • These solutions provide plenty of questions to practice.

NCERT solutions of class 12 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Notes of class 12 - Subject Wise

Given below are the subject-wise NCERT Notes of class 12 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 12:

NCERT Exemplar Class 12 Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 12 NCERT:


Frequently Asked Questions (FAQs)

1. What are the important topics of this chapter?

Introduction to Inverse Trigonometric Functions, The Basic Concepts of Inverse Trigonometric Functions and Properties of Inverse Trigonometric Functions are important topics of this chapter.

2. Are these solutions helpful for board examinations?

Yes, the NCERT exemplar Class 12 Maths chapter 2 solutions are helpful for you to prepare for board exams.

3. How many questions are there in this chapter?

There is only 1 exercise in this chapter with 55 problem solving questions.

4. Are these solutions helpful for competitive examinations?

Yes, NCERT exemplar solutions for Class 12 Maths chapter 2 cover syllabus for exams are very reliable for preparing for competitive entrance exams like NEET and JEE Main.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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