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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
Upcoming Event
CBSE Class 12th  Exam Date : 01 Jan' 2025 - 14 Feb' 2025

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.

This Story also Contains
  1. NCERT Exemplar Class 12 Maths Solutions Chapter 2: Exercise-1.3
  2. Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions
  3. Sub-Topics Covered in NCERT Exemplar Class 12 Maths Solutions Chapter 2
  4. What student will learn in NCERT Exemplar Class 12 Maths Solutions Chapter 2?
  5. NCERT Exemplar Class 12 Maths Solutions
  6. Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 2

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.

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:19

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

Answer:

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

Question:20

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

Answer:

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

Question:21

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

Answer:

Answer :(D)

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

Question:22

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

Answer:

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

Question:23

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

Answer:

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

Question:24

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

Answer:

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

Question:25

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

Answer:

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

Question:26

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

Answer:

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

Question:27

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

Answer:

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

Question:28

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

Answer:

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

Question:29

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

Answer:

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

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

Question:30

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

Answer:

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

Question:31

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

Answer:

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

Question:32

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

Answer:

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

Question:33

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

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

Question:34

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

Answer:

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

Question:35

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

Answer:

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

Question:36

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

Answer:

Answer:(A)
We have , \sqrt{1+\cos 2x}=\sqrt{2}\cos^{-1}\left ( \cos x \right ), x is in \left [ \frac{\pi}{2}, \pi \right ]
R.H.S
\sqrt{2}\cos^{-1}\left ( \cos x \right )=\sqrt{2}x
So, \sqrt{1+\cos 2x}=\sqrt{2}x
Squaring both side , we get,
\left ( 1+\cos 2 x \right )=2x^{2}
\Rightarrow \cos 2x=2x^{2}-1
Now plotting cos 2x and 2x2-1, we get,
a36
As , there is no point of intersection in \left [ \frac{\pi}{2},\pi \right ], so therre is no
solution of the given equation in \left [ \frac{\pi}{2},\pi \right ]

Question:37

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

Answer:

Answer :(C)
Plotting cos-1 x and sin-1 x, we get,
a37
As, graph of cos-1 x is above graph of sin-1 x in \left [ -1,\frac{1}{\sqrt{2}} \right ).
So, cos–1x > sin–1 x for all x in \left [ -1,\frac{1}{\sqrt{2}} \right ) .

Question:38

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

Answer:

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

Question:39

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

Answer:

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

Question:40

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

Answer:

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

Question:40

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

Answer:

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

Question:41

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

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

Question:42

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

Answer:

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

Question:43

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

Answer:

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

Question:44

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

Answer:

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

Question:45

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

Answer:

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

Question:46

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

Answer:

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

Question:47

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

Answer:

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

Question:48

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

Answer:

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

Question:49

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

Answer:

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

Question:50

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

Answer:

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

Question:51

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

Answer:

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

Question:52

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

Answer:

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

Question:53

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

Answer:

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

Question:54

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

Answer:

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

Question:55

State True or False for the statement

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

Answer:

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

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.

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 Questions (FAQs)

1. What are the important topics of this chapter?

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

2. Are these solutions helpful for board examinations?

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

3. How many questions are there in this chapter?

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

4. Are these solutions helpful for competitive examinations?

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

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
Can I change my board from CBSE to CHSE Odisha in Class 12th?

Hello,

Yes, you can switch from CBSE to CHSE Odisha in Class 12th, but there are a few conditions:

  1. Eligibility : Check if CHSE Odisha allows lateral entry into Class 12th. Generally, students can transfer from CBSE to state boards with prior approval.
  2. Document Requirements : You may need to provide transcripts, mark sheets, and a transfer certificate.
  3. Subject Alignment : Ensure that the subjects you studied in CBSE align with the CHSE curriculum.
  4. Application Process : Contact the respective CHSE Odisha authority or school for specific transfer procedures and deadlines.

Hope it helps !

Read Complete Answer
Prachi Kumari 30 December,2024
quartely question paper for mathematics cbse

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!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

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.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
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Popular CBSE Class 12th Questions

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
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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

 
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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
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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
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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 .
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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
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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.
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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.
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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
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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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