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NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2 - Inverse Trigonometric Functions

NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2 - Inverse Trigonometric Functions

Edited By Ramraj Saini | Updated on Dec 03, 2023 01:48 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 2 Exercise 2.2

NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2 Inverse Trigonometric Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT Solutions for Class 12 Maths chapter 2 exercise 2.1 talks about the principal values in a given range of various trigonometric functions. Exercise 2.2 Class 12 Maths basically deals with questions in which a certain range of the angle is provided. NCERT Solutions for Class 12 Maths chapter 2 exercise 2.2 has some questions which are quite tricky. Silly mistake chances are quite high. So practice them well before the exam with precision. Also below is the list of other NCERT exercises which can be referred to for more understanding.

This Story also Contains
  1. NCERT Solutions For Class 12 Maths Chapter 2 Exercise 2.2
  2. Assess NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2
  3. NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Exercise 2.2
  4. Question:1 Prove the following:
  5. More About NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2
  6. Benefits of NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2
  7. Key Features Of NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2
  8. Subject wise NCERT Exemplar solutions

12th class Maths exercise 2.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

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NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Exercise 2.2

Question:1 Prove the following: 3\sin^{-1}x = \sin^{-1}(3x - 4x^3),\;\;x\in\left[-\frac{1}{2},\frac{1}{2} \right ]

Answer:

Given to prove: 3\sin^{-1}x = \sin^{-1}(3x - 4x^3)

where, x\:\epsilon \left[-\frac{1}{2},\frac{1}{2} \right ] .

Take \theta= \sin ^{-1}x or x = \sin \theta

Take R.H.S value

\sin^{-1}(3x - 4x^3)

= \sin^{-1}(3\sin \theta - 4\sin^3 \theta)

= \sin^{-1}(\sin 3\theta)

= 3\theta

= 3\sin^{-1}x = L.H.S

Question:2 Prove the following: 3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ]

Answer:

Given to prove 3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ] .

Take \cos^{-1}x = \theta or \cos \theta = x ;

Then we have;

R.H.S.

\cos^{-1}(4x^3 - 3x)

= \cos^{-1}(4\cos^3 \theta - 3\cos\theta) \left [ \because 4\cos^3 \theta - 3\cos\theta = \cos3 \theta \right ]

= \cos^{-1}(\cos3\theta)

= 3\theta

= 3\cos^{-1}x = L.H.S

Hence Proved.

Question:3 Prove the following: \tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} = \tan^{-1}\frac{1}{2}

Answer:

Given to prove \tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} = \tan^{-1}\frac{1}{2}

We have L.H.S

\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24}

=\tan^{-1}\frac{\frac{2}{11} + \frac{7}{24} }{1 - \left ( \frac{2}{11}\times\frac{7}{24} \right ) } \left [ \because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x +y}{1 - xy} \right ]

=\tan^{-1}\frac{11\times 24 }{\frac{11\times24 -14}{11\times 24} }

=\tan^{-1}\frac{48 + 77}{264 -14}

=\tan^{-1}\left ( \frac{125}{250}\right ) = \tan^{-1}\left ( \frac{1}{2} \right )

= R.H.S

Hence proved.

Question:4 Prove the following: 2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}

Answer:

Given to prove 2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}

Then taking L.H.S.

We have 2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7}

=\tan^{-1} \frac{2.\frac{1}{2}}{1 - \left ( \frac{1}{2} \right )^2} + \tan^{-1} \frac{1}{7} \because 2\tan^{-1} x = \tan^{-1} \frac{2x}{1- x^2}

=\tan^{-1} \frac{1}{(\frac{3}{4})} + \tan^{-1} \frac{1}{7}

=\tan^{-1} \frac{4}{3} + \tan^{-1} \frac{1}{7}

=\tan^{-1} \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3}.\frac{1}{7}} \left [ \because \tan^{-1}x + \tan^{-1} y = \tan^{-1} \frac{x +y}{1- xy}\right ]

=\tan^{-1} \left ( \frac{\frac{28+3}{21}}{\frac{21-4}{21}} \right )

=\tan^{-1} \frac{31}{17}

= R.H.S.

Hence proved.

Question:5 Write the following functions in the simplest form: \tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x},\;\;x\neq 0

Answer:

We have \tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x}

Take

\therefore \tan^{-1} \frac {\sqrt{1+x^2} - 1}{x} = \tan^{-1}\frac{\sqrt{1+\tan^2 \Theta - 1}}{\tan \Theta}

=\tan^{-1}(\frac{sec \Theta-1}{tan \Theta}) = \tan^{-1}\left ( \frac{1-cos \Theta}{sin \Theta} \right )

=\tan^{-1}\left ( \frac {2sin^2\left ( \frac{\Theta}{2} \right )}{2sin\frac{\Theta}{2}cos\frac{\Theta}{2}} \right )

=\tan^{-1}\left ( \tan\frac{\Theta}{2} \right ) = \frac{\Theta}{2} =\frac{1}{2}\tan^{-1}x

=\frac{1}{2}\tan^{-1}x is the simplified form.

Question:6 Write the following functions in the simplest form : \tan^{-1} \frac{1}{\sqrt{x^2 -1}},\;\; |x| > 1

Answer:

Given that \tan^{-1} \frac{1}{\sqrt{x^2 -1}},\;\; |x| > 1

Take x =cosec\ \Theta or \Theta = cosec ^{-1}x

\therefore tan^{-1}\frac{1}{\sqrt{x^2-1}}

=tan^{-1} \frac{1}{\sqrt{cosec^2 \Theta -1}}

=tan^{-1}(\frac{1}{\cot \Theta})

=tan^{-1}(\tan \Theta) = \Theta

= cosec^{-1}x

=\frac{\pi}{2}- \sec^{-1}x [\because cosec^{-1}x + \sec^{-1}x = \frac{\pi}{2}]

Question:7 Write the following functions in the simplest form: \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi

Answer:

Given that \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi

We have in inside the root the term : \frac{1-\cos x}{1 + \cos x}

Put 1-\cos x = 2\sin^2\frac{x}{2} and 1+\cos x = 2\cos^2\frac{x}{2} ,

Then we have,

=\tan^{-1}\left(\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}} \right )

=\tan^{-1}\left( \frac{\sin \frac{x}{2}}{\cos\frac{x}{2}} \right )

=\tan^{-1}(\tan\frac{x}{2}) = \frac{x}{2}

Hence the simplest form is \frac{x}{2}

Question:8 Write the following functions in the simplest form: \tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right ),\;\; \frac{-\pi}{4} < x < \frac{3\pi}{4}

Answer:

Given \tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right ) where x\:\epsilon\:( \frac{-\pi}{4} < x < \frac{3\pi}{4})

So,

=\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right )

Taking \cos x common from numerator and denominator.

We get:

=\tan^{-1}\left(\frac{1 -(\frac{\sin x}{\cos x}) }{1+(\frac{\sin x}{\cos x}) } \right )

=\tan^{-1}\left(\frac{1 - \tan x }{1+\tan x } \right )

= \tan^{-1}(1) - \tan^{-1}(\tan x) as, \left [ \because \tan^{-1}x - \tan^{-1}y = \frac{x - y}{1 + xy} \right ]

= \frac{\pi}{4} - x is the simplest form.

Question:9 Write the following functions in the simplest form: \tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a

Answer:

Given that \tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a

Take x = a\sin \theta or

\theta = \sin^{-1}\left ( \frac{x}{a} \right ) and putting it in the equation above;

\tan^{-1} \frac{a\sin \theta}{\sqrt{a^2 - (a\sin \theta)^2}}

=\tan^{-1} \frac{a\sin \theta}{a\sqrt{1 - \sin^2 \theta}}

=\tan^{-1} \left ( \frac{\sin \theta}{\sqrt{\cos^2 \theta}} \right ) = \tan^{-1} \left ( \frac{\sin \theta}{{\cos \theta}} \right )

=\tan^{-1}\left ( \tan \theta \right )

=\theta = \sin^{-1}\left ( \frac{x}{a} \right ) is the simplest form.

Question:10 Write the following functions in the simplest form: \tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right ),\;\;a>0\;\;;\;\;\frac{-a}{\sqrt3} < x < \frac{a}{\sqrt3}

Answer:

Given \tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )

Here we can take x = a\tan \theta \Rightarrow \frac{x}{a} = \tan \theta

So, \theta = \tan^{-1}\left ( \frac{x}{a} \right )

\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right ) will become;

=\tan^{-1}\left(\frac{3a^2a\tan \theta -(a\tan \theta)^3}{a^3 - 3a(a\tan \theta)^2} \right ) = \tan^{-1}\left(\frac{3a^3\tan \theta -a^3\tan ^3 \theta}{a^3 - 3a^3\tan ^2 \theta} \right )

and as \left [ \because \left(\frac{3\tan \theta -\tan ^3 \theta}{ 1- 3\tan ^2 \theta} \right) =\tan 3\theta \right ] ;

=3 \theta

=3 \tan^{-1}(\frac{x}{a})

hence the simplest form is 3 \tan^{-1}(\frac{x}{a}) .

Question:11 Find the values of each of the following: \tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]

Answer:

Given equation:

\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]

So, solving the inner bracket first, we take the value of \sin x^{-1} \frac{1}{2} = x.

Then we have,

\sin x = \frac{1}{2} = \sin \left ( \frac{\pi}{6} \right )

Therefore, we can write \sin^{-1} \frac{1}{2} = \frac{\pi}{6} .

\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ] = \tan^{-1}\left[2\cos\left(2\times\frac{\pi}{6} \right ) \right ]

= \tan^{-1}\left[2\cos\left(\frac{\pi}{3} \right ) \right ] = \tan^{-1}\left[2\times\left(\frac{1}{2} \right ) \right ] = \tan^{-1}1 = \frac{\pi}{4} .

Question:12 Find the values of each of the following: \cot(\tan^{-1}a + \cot^{-1}a)

Answer:

We have to find the value of \cot(\tan^{-1}a + \cot^{-1}a)

As we know \left [\because \tan^{-1}x + \cot^{-1} x = \frac{\pi}{2} \right ] so,

Equation reduces to \cot(\frac{\pi}{2}) = 0 .

Question:13 Find the values of each of the following: \tan \frac{1}{2}\left[\sin^{-1}\frac{2x}{1+x^2} + cos^{-1}\frac{1-y^2}{1+y^2} \right ],\;\;|x|<1,\;y>0 and xy<1

Answer:

Taking the value x = \tan \Theta or \tan^{-1}x = \Theta and y = \tan \Theta or \tan^{-1} y = \Theta then we have,

= \tan \frac{1}{2}\left[\sin^{-1}\frac{2\tan \Theta}{1+(\tan \Theta)^2} + cos^{-1}\frac{1-\tan^2 \Theta}{1+(\tan \Theta)^2} \right ] ,

= \tan \frac{1}{2}\left[\sin^{-1}(\sin2\Theta) + cos^{-1} (\cos 2\Theta) \right ]

\because \left[\cos^{-1}(\frac{1-\tan^2 \Theta}{1+ \tan^2\Theta}) = \cos^{-1} (\cos2 \Theta) , \right ]

\because \left[\sin^{-1}(\frac{2\tan\Theta}{1+ \tan^2\Theta}) = \sin^{-1} (\sin2 \Theta) \right ]

Then,

=\tan \frac{1}{2}\left[2\tan^{-1}x + 2\tan^{-1}y \right ] \because \left[\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right]

=\tan \left [ \tan^{-1}\frac{x+y}{1-xy} \right ]

=\frac{x+y}{1-xy} Ans.

Question:14 If \sin\left(\sin^{-1}\frac{1}{5} + \cos ^{-1}x \right ) =1 , then find the value of x .

Answer:

As we know the identity;

sin^{-1} x + cos^{-1} x = \frac {\pi}{2},\ x\ \epsilon\ [-1,1] . it will just hit you by practice to apply this.

So, \sin\left(\sin^{-1}\frac{1}{5} + \cos ^{-1}x \right ) =1 or \sin^{-1}\frac{1}{5} + \cos ^{-1}x =\sin^{-1}(1) ,

we can then write \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x ,

putting in above equation we get;

\sin^{-1}\frac{1}{5} + \frac{\pi}{2} - \sin^{-1}x =\frac{\pi}{2} \because \left [ \sin^{-1}(1)=\frac{\pi}{2} \right ]

= \sin^{-1}x = \sin^{-1} \frac{1}{5}

Ans. x = \frac{1}{5}

Question:15 If \tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} =\frac{\pi}{4} , then find the value of x .

Answer:

Using the identity \tan^{-1}x+\tan^{-1} y = \tan^{-1}{\frac{x+y}{1-xy}} ,

We can find the value of x;

So, \tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} =\frac{\pi}{4}

on applying,

= \tan^{-1}{\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1- \left ( \frac{x-1}{x-2} \right )\left ( \frac{x+1}{x+2} \right )}}

=\tan^{-1}\frac{\frac{(x-1)(x+2)+(x-2)(x+1)}{x^2-4}}{1-\frac{x^2-1}{x^2-4}} = \tan^{-1} \left [ \frac{2x^2-4}{-3} \right ] = \frac{\pi}{4}

=\frac{2x^2-4}{-3} = \tan (\frac{\pi}{4})=1

= 2x^2=1 or x = \pm \frac{1}{\sqrt{2}} ,

Hence, the possible values of x are \pm \frac{1}{\sqrt{2}} .

Question:16 Find the values of each of the expressions in Exercises 16 to 18. \sin^{-1}\left (\sin\frac{2\pi}{3} \right )

Answer:

Given \sin^{-1}\left (\sin\frac{2\pi}{3} \right ) ;

We know that \sin^{-1}(\sin x) = x

If the value of x belongs to \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] then we get the principal values of \sin^{-1}x .

Here, \frac{2\pi}{3} \notin \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]

We can write \sin^{-1}\left (\sin\frac{2\pi}{3} \right ) is as:

= \sin^{-1}\left [ \sin\left ( \pi-\frac{2\pi}{3} \right ) \right ]

= \sin^{-1}\left [ \sin \frac{\pi}{3} \right ] where \frac{\pi}{3} \epsilon \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]

\therefore \sin^{-1}\left (\sin\frac{2\pi}{3} \right )=\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]=\frac{\pi}{3}

Question:17 Find the values of each of the expressions in Exercises 16 to 18. \tan^{-1}\left (\tan\frac{3\pi}{4} \right )

Answer:

As we know \tan^{-1}\left ( \tan x \right ) =x

If x \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ). which is the principal value range of \tan^{-1}x .

So, as in \tan^{-1}\left (\tan\frac{3\pi}{4} \right ) ;

\frac{3\pi}{4}\notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )

Hence we can write \tan^{-1}\left (\tan\frac{3\pi}{4} \right ) as :

\tan^{-1}\left (\tan\frac{3\pi}{4} \right ) = \tan^{-1}\left (\tan\frac{3\pi}{4} \right) = \tan^{-1}\left [ \tan(\pi - \frac{\pi}{4}) \right ] = \tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]

Where -\frac{\pi}{4} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )

and \therefore \tan^{-1}\left (\tan\frac{3\pi}{4} \right )=\tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]=-\frac{\pi}{4}

Question:18 Find the values of each of the expressions in Exercises 16 to 18. \tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )

Answer:

Given that \tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )

we can take \sin^{-1}\frac{3}{5} = x ,

then \sin x = \frac{3}{5}

or \cos x = \sqrt{1-\sin^{2}x}= \frac{4}{5}

\Rightarrow \tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}

\Rightarrow \tan^{-1}\frac{3}{4}= x

We have similarly;

\cot^{-1} \frac{3}{2} = \tan^{-1} \frac{2}{3}

Therefore we can write \tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )

=\tan\left(\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{2}{3} \right )

=\tan\left[\tan^{-1}\left ( \frac{\frac{3}{4}+\frac{2}{3}}{1- \frac{3}{4}.\frac{2}{3}} \right ) \right ] from As, \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]

=\tan \left (\tan^{-1} \frac{9+8}{12-6} \right ) = \tan \left (\tan^{-1} \frac{17}{6} \right )= \frac{17}{6}

Question:19 \cos^{-1}\left(\cos\frac{7\pi}{6} \right ) is equal to

(A) \frac{7\pi}{6}

(B) \frac{5\pi}{6}

(C) \frac{\pi}{3}

(D) \frac{\pi}{6}

Answer:

As we know that \cos^{-1} (cos x ) = x if x\epsilon [0,\pi] and is principal value range of \cos^{-1}x .

In this case \cos^{-1}\left(\cos\frac{7\pi}{6} \right ) ,

\frac{7\pi}{6} \notin [0,\pi]

hence we have then,

\cos^{-1}\left(\cos\frac{7\pi}{6} \right ) = \cos^{-1} \left ( \cos \frac{-7\pi}{6} \right ) = \cos^{-1}\left [ \cos\left ( 2\pi - \frac{7\pi}{6} \right ) \right ]

\left [ \because \cos (2\pi + x) = \cos x \right ]

\therefore\ we\ have \cos^{-1}\left ( \cos \frac{7\pi}{6} \right ) = \cos^{-1}\left ( \cos \frac{5\pi}{6} \right ) = \frac{5\pi}{6}

Hence the correct answer is \frac{5\pi}{6} (B).

Question:20 \sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) is equal to

(A) \frac{1}{2}

(B)

(C) \frac{1}{4}

(D) 1

Answer:

Solving the inner bracket of \sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) ;

\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) or

Take \sin^{-1}\left(-\frac{1}{2} \right ) = x then,

\sin x =-\frac{1}{2} and we know the range of principal value of \sin^{-1}x\ is\ \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ].

Therefore we have \sin^{-1}\left ( -\frac{1}{2} \right ) = -\frac{\pi}{6} .

Hence, \sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) = \sin \left ( \frac{\pi}{3}+ \frac{\pi}{6} \right )= \sin \left ( \frac{3\pi}{6} \right ) = \sin\left ( \frac{\pi}{2} \right ) = 1

Hence the correct answer is D.

Question:21 \tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3) is equal to

(A) \pi

(B) -\frac{\pi}{2}

(C) 0

(D) 2\sqrt3

Answer:

We have \tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3) ;

finding the value of \cot^{-1}(-\sqrt3) :

Assume \cot^{-1}(-\sqrt3) =y then,

\cot y = -\sqrt 3 and the range of the principal value of \cot^{-1} is (0,\pi) .

Hence, principal value is \frac{5\pi}{6}

Therefore \cot^{-1} (-\sqrt3) = \frac {5\pi}{6}

and \tan^{-1} \sqrt3 = \frac{\pi}{3}

so, we have now,

\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)=\frac{\pi}{3} - \frac{5\pi}{6}

= \frac{2\pi - 5\pi}{6} = \frac{-3\pi}{6}

or, = \frac{ -\pi}{2}

Hence the answer is option (B).

More About NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

The NCERT class 12 maths chapter Inverse Trigonometric Functions deals with questions taking from basic to advanced level. Exercise 2.2 Class 12 Maths has some moderate level of questions which are important for the examination. NCERT Solutions for class 12 maths chapter 2 exercise 2.2 along with NCERT exemplar questions is sufficient for a good understanding.

Also Read| Inverse Trigonometric Functions NCERT Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

  • The Class 12th maths chapter 2 exercise is described in a very easy manner. Students can comprehend by reading these notes.
  • Exercise 2.2 Class 12 Maths is the extension of NCERT syllabus Exercise 2.1, as it has some difficult questions.
  • NCERT book Class 12 Maths chapter 2 exercise 2.2 solutions has some questions which can take time, but one should make sure to complete them for better understanding.
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Key Features Of NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 2.2 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 2.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 2.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 2.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 2.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 2.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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Frequently Asked Question (FAQs)

1. Is inverse trigonometric functions used in Physics also ?

Yes, just like integration, it is quite useful in Physics as well as Chemistry. These concepts are discussed in ex 2.2 class 12 comprehensively. To find the value of angles its quite useful. NCERT syllabus can be followed for the preparation of CBSE board exam.

2. Inverse functions of the trigonometric functions are known as ……..?

Inverse trigonometric functions

3. What is the use of Inverse trigonometric functions?

It is used to find angles from a given angle’s trigonometric ratios. 

4. How to memorise principle values of basic inverse trigonometric functions ?

Make short notes and revise them multiple times. Practice questions so the brain retains it. For more questions use NCERT exemplar.

5. Is it required to memorise the principal value of basic inverse trigonometric functions ?

No, It is not mandatory but solving questions becomes easy if values are by heart. 

6. How to solve proof related questions ?

Take step by step method to reach what is asked starting from the given conditions.

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Related E-books & Sample Papers

CBSE 12th Hindi B syllabus 2024-25

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
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
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

Read Complete Answer
satyamercedes0626 23 July,2024
I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.


Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.


Hope this resolves your query.

Read Complete Answer
Abhishek 12 July,2024
View All

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
Read More

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

 
Read More

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
Read More

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
Read More

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 .
Read More

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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

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