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NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

Edited By Ravindra Pindel | Updated on Sep 15, 2022 01:55 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 2 contains the topic that requires some time and effort on the student’s part. It is an important chapter in Calculus Mathematics, which helps in understating the integrals and their existence. We will help you to find the solutions of all the NCERT exemplar questions of this chapter. Through NCERT Exemplar Class 12 Math chapter 2 solutions, the students will learn about inverse trigonometric functions. Students looking for NCERT exemplar Class 12 Maths solutions chapter 2 PDF download can use the online webpage to PDF tool.

More about NCERT exemplar Class 12 Maths solutions chapter 2 Inverse Trigonometric Functions

In NCERT exemplar solutions for Class 12 Maths chapter 2, one will learn about how to find integrals and what are the ranges and domains of these inverses functions . Some other concepts including what makes these functions inverse, how these functions behave and Various properties are explained in NCERT exemplar Class 12 Maths chapter 2 solutions. Thus, this chapter of NCERT Class 12 Maths Solutions will help the students to solve the questions at a later stage of the chapter.

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NCERT Exemplar Class 12 Maths Solutions Chapter 2: Exercise-1.3

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: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:

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}

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

  • Class 12 Math NCERT Exemplar solutions chapter 2 holds a lot of importance for those who want to pursue a future in engineering and science.
  • Understanding sine, cosine, tangent and their inverse trigonometric functions along with their inverse, is crucial.
  • We will help in understanding the topics much easily by solving the questions in a way that is simple and exhaustive.
  • Our highly experienced guides and teachers have solved the questions in the simplest language possible. Solutions will make it easier for the students to understand the topic and the questions.
  • NCERT Exemplar solutions for Class 12 Math chapter 2 are detailed with additional extra steps with formulas at every step. This will make the answer detailed for each student out there.
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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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What student will learn in NCERT Exemplar Class 12 Maths Solutions Chapter 2?

  • With our exhaustive 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.
  • Several topics are covered in this chapter that will help you get prepared for higher education and the exams likewise. NCERT Exemplar Class 12 Math solutions chapter 2 covers inverse trigonometric functions, their principles, range, domain and functions.
  • Topics given in NCERT exemplar Class 12 Maths solutions chapter 2 will help in understanding the inverse trigonometric functions in a much more detailed way with the help of questions and solved examples.
  • To solve questions and to understand theorems and rules, it is crucial to understand the properties of each and every inverse trigonometric function. Therefore, the topic covers the relations and properties of the cot, tan, sine, sec, cosec, cos, and sec inverse trigonometric functions.
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NCERT Exemplar Class 12 Maths Solutions

Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 2

  • In NCERT Exemplar solutions for Class 12 Math chapter 2, students should 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.
  • In the Class 12 Maths NCERT exemplar solutions chapter 2, the student will learn in detail about basic concepts of the trigonometric functions like sine, cosine, cosec, functions and their features.

NCERT Exemplar Class 12 Solutions

Also, check NCERT Solutions for questions given in the book:

Chapter 1

Relations and Functions

Chapter 2

Inverse Trigonometric Functions

Chapter 3

Matrices

Chapter 4

Determinants

Chapter 5

Continuity and Differentiability

Chapter 6

Application of Derivatives

Chapter 7

Integrals

Chapter 8

Application of Integrals

Chapter 9

Differential Equations

Chapter 10

Vector Algebra

Chapter 11

Three Dimensional Geometry

Chapter 12

Linear Programming

Chapter 13

Probability

Must Read NCERT Solution subject wise

Read more NCERT Notes subject wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (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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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


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Choosing Your Stream:

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Engineering Subjects (for a Degree):

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  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
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Looking Ahead (2025 Admissions):

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  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

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However, the shortlisting process is multifaceted:

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Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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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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