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

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.

  • Inverse Trigonometric Functions Exercise 2.1

  • Inverse Trigonometric Functions Miscellaneous Exercise

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.

Also see-

  • NCERT exemplar solutions class 12 maths chapter 2

  • NCERT solutions for class 12 maths chapter 2

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Subject wise NCERT Exemplar solutions

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