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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.
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.
Class 12 Maths Chapter 2 exemplar solutions Exercise: 2.3 Page number: 35-41 Total questions: 55 |
Question:1
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
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
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
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
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
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
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
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
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
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,
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
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
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
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
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
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
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
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
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,
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \sin \left [ \frac{\pi}{3} \right ]\right ]$
$\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\frac{\pi}{3}$
The sub-topics that are covered in this chapter of inverse trigonometric functions are:
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.
Here are the subject-wise links for the NCERT solutions of class 12:
Given below are the subject-wise NCERT Notes of class 12 :
Here are some useful links for NCERT books and the NCERT syllabus for class 12:
Given below are the subject-wise exemplar solutions of class 12 NCERT:
Introduction to Inverse Trigonometric Functions, The Basic Concepts of Inverse Trigonometric Functions and Properties of Inverse Trigonometric Functions are important topics of this chapter.
Yes, the NCERT exemplar Class 12 Maths chapter 2 solutions are helpful for you to prepare for board exams.
There is only 1 exercise in this chapter with 55 problem solving questions.
Application Date:24 March,2025 - 23 April,2025
Admit Card Date:04 April,2025 - 26 April,2025
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
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I hope this information helps you.
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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
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