NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

 

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Two chapters 'relation and function' and 'inverse trigonometry' of NCERT Class 12 has 10 % weightage in the board examination. In this article, you will find NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions which will help you to understand the concepts in a detailed manner. In class 11 maths you have already learnt about trigonometric functions. It won't take much effort to command on inverse trigonometric functions if you have good knowledge of trigonometric functions. You just need to practice NCERT questions including examples and miscellaneous exercise. You may find some difficulties in solving the problems, so you can take the help of these solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions. These solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions are very important for the board exam and as well as for competitive exams like JEE Main, BITSAT, VITEEE, etc. In this chapter, there are 2 exercises with 35 questions. All these questions are prepared and explained in a step-by-step method in the NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions article.  So, it will be very easy for you to understand the concept. Check all  NCERT solutions from class 6 to class 12 to get a better understanding of the concepts.

There are important applications of ITF in geometry, navigation, science, and engineering. Also, inverse trigonometric functions play an important role in the calculus part of mathematics to define many integrals. Many students have a misconception in class 12 maths chapter inverse trigonometric functions like\sin^-^1x=\frac{1}{\sin x}  . But the inverse function (f^-^1) is not the same as  \frac{1}{f} for example \sin^-^1x\neq \frac{1}{\sin x}.

For inverse to exist, the function must be one-one and onto but trigonometric functions are neither one-one and onto over their domain and natural ranges. So, to ensure the existence of their inverse we restrict domains and ranges of trigonometric functions. And this range is known as principal value. In the following table the principal value branches of inverse trigonometric functions(ITF) are given:

                                                

Topics of NCERT Grade 12 Maths Chapter-2 Inverse Trigonometric Functions

2.1 Introduction

2.2 Basic Concepts

2.3 Properties of Inverse Trigonometric Functions

 

CBSE NCERT Solutions for class 12 maths chapter 2 Inverse Trigonometric Functions: Exercise 2.1

Question:1 Find the principal values of the following : \sin^{-1}\left ( \frac{-1}{2} \right )

Answer:

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

\implies \sin x = \frac{-1}{2}= -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})
We know, principle value range of sin^{-1} is [-\frac{\pi}{2}, \frac{\pi}{2}]

\therefore The principal value of \sin^{-1}\left ( \frac{-1}{2} \right ) is -\frac{\pi}{6},

Question:2 Find the principal values of the following: \cos^{-1}\left(\frac{\sqrt3}{2} \right )

Answer:

So, let us assume that \cos^{-1}\left(\frac{\sqrt3}{2} \right ) = x  then,

Taking inverse both sides we get;

cos\ x = (\frac{\sqrt{3}}{2}),  or   cos (\frac{\pi}{6}) = (\frac{\sqrt{3}}{2})         

and as we know that the principal values of cos^{-1} is from [0,\pi],

Hence cos\ x = (\frac{\sqrt{3}}{2})     when x = \frac{\pi}{6} . 

Therefore, the principal value for \cos^{-1}\left(\frac{\sqrt3}{2} \right )is   \frac{\pi}{6}.

Question:3 Find the principal values of the following: \textup{cosec}^{-1}(2)

Answer:

Let us assume that \textup{cosec}^{-1}(2) = x, then we have;

Cosec\ x = 2, or 

Cosec( \frac{\pi}{6}) = 2  .

And we know the range of principal values is  [\frac{-\pi}{2},\frac{\pi}{2}] - \left \{ 0 \right \}.

Therefore the principal value of  \textup{cosec}^{-1}(2) is \frac{\pi}{6}.

Question:4 Find the principal values of the following: \tan^{-1}(-\sqrt3)

Answer:

Let us assume that \tan^{-1}(-\sqrt3) = x, then we have;

\tan x = (-\sqrt 3) or

 -\tan (\frac{\pi}{3}) = \tan \left ( \frac{-\pi}{3} \right ).

and as we know that the principal value of \tan^{-1} is \left ( \frac{-\pi}{2}, \frac{\pi}{2} \right ).

Hence the only principal value of \tan^{-1}(-\sqrt3) when x = \frac{-\pi}{3}.

Question:5 Find the principal values of the following: \cos^{-1}\left(-\frac{1}{2} \right )

Answer:

Let us assume that  \cos^{-1}\left(-\frac{1}{2} \right ) =y then,

Easily we have; \cos y = \left ( \frac{-1}{2} \right ) or we can write it as: 

-\cos \left ( \frac{\pi}{3} \right ) = \cos \left ( \pi - \frac{\pi}{3} \right ) = \cos \left ( \frac{2\pi}{3} \right ).

as we know that the range of the principal values of \cos^{-1}  is \left [ 0,\pi \right ].

Hence \frac{2\pi}{3} lies in the range it is a principal solution.

Question:6 Find the principal values of the following: \tan^{-1}(-1)

Answer:

Given \tan^{-1}(-1) so we can assume it to be equal to 'z';

\tan^{-1}(-1) =z  , 

\tan z = -1     

or        

-\tan (\frac{\pi}{4}) = \tan(\frac{-\pi}{4})= -1                                                 

 And as we know the range of principal values of \tan^{-1} from \left ( \frac{-\pi}{2}, \frac{\pi}{2} \right ).

As only one value z = -\frac{\pi}{4} lies hence we have only one principal value that is -\frac{\pi}{4}.

Question:7 Find the principal values of the following: \sec^{-1}\left (\frac{2}{\sqrt3}\right)

Answer:

Let us assume that \sec^{-1}\left (\frac{2}{\sqrt3}\right) = z  then,

we can also write it as; \sec z = \left (\frac{2}{\sqrt3}\right).

Or \sec (\frac{\pi}{6}) = \left (\frac{2}{\sqrt3}\right)  and the principal values lies between  \left [ 0, \pi \right ] - \left \{ \frac{\pi}{2} \right \}.

Hence we get only one principal value of \sec^{-1}\left (\frac{2}{\sqrt3}\right)  i.e., \frac{\pi}{6}.

Question:8 Find the principal values of the following: \cot^{-1}(\sqrt3)

Answer:

Let us assume that \cot^{-1}(\sqrt3) = x , then we can write in other way,

\cot x = (\sqrt3)   or   

\cot (\frac{\pi}{6}) = (\sqrt3).

Hence when   x=\frac{\pi}{6}  we have \cot (\frac{\pi}{6}) = (\sqrt3).

and the range of principal values of \cot^{-1} lies in \left ( 0, \pi \right ).

Then the principal value of \cot^{-1}(\sqrt3) is \frac{\pi}{6}

Question:9 Find the principal values of the following: \cos^{-1}\left(-\frac{1}{\sqrt2} \right )

Answer:

Let us assume \cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x;

Then we have \cos x = \left ( \frac{-1}{\sqrt 2} \right )   

or

   -\cos (\frac{\pi}{4}) = \left ( \frac{-1}{\sqrt 2} \right )  ,

\cos (\pi - \frac{\pi}{4}) = \cos (\frac{3\pi}{4}).

And we know the range of principal values of \cos^{-1} is [0,\pi].

So, the only principal value which satisfies \cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x is \frac{3\pi}{4}.

Question:10 Find the principal values of the following: \textup{cosec}^{-1}(-\sqrt2)

Answer:

Let us assume the value of \textup{cosec}^{-1}(-\sqrt2) = y , then

we have cosec\ y = (-\sqrt 2)   or

 -cosec\ (\frac{\pi}{4}) = (-\sqrt 2) = cosec\ (\frac{-\pi}{4}).

and the range of the principal values of \textup{cosec}^{-1}  lies between  \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] - \left \{ 0 \right \}.

hence the principal value of \textup{cosec}^{-1}(-\sqrt2) is \frac{-\pi}{4}.

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

Answer:

To find the values first we declare each term to some constant ;

tan^{-1}(1) = x , So we have \tan x = 1;

or \tan (\frac{\pi}{4}) = 1 

Therefore, x = \frac{\pi}{4} 

 cos^{-1}(\frac{-1}{2}) = y 

So, we have

 \cos y = \left ( \frac{-1}{2} \right ) = -\cos \left ( \frac{\pi}{3} \right ) = \cos(\pi - \frac{\pi}{3}) = \cos \left ( \frac{2\pi}{3} \right ).

Therefore y = \frac{2\pi}{3},

\sin^{-1}(\frac{-1}{2}) = z,

So we have;

\sin z = \frac{-1}{2}   or  -\sin (\frac{\pi}{6}) =\sin (\frac{-\pi}{6}) = \frac{-1}{2}

Therefore z = -\frac{\pi}{6}

Hence we can calculate the sum:

= \frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}

=\frac{3\pi + 8\pi -2\pi}{12} = \frac{9\pi}{12}=\frac{3\pi}{4}.

Question:12 Find the values of the following: \cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )

Answer:

Here we have \cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )

let us assume that the value of

\cos^{-1}\left ( \frac{1}{2} \right ) = x, \:and\:\sin^{-1}\left(\frac{1}{2} \right ) = y;

then we have to find out the value of x +2y.

Calculation of x :

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

 \Rightarrow \cos x = \frac{1}{2}   

    \Rightarrow \cos \frac{\pi}{3} = \frac{1}{2}  , 

Hence x = \frac{\pi}{3}.

Calculation of y :

\Rightarrow \sin^{-1}\left(\frac{1}{2} \right ) = y  

 \Rightarrow \sin y = \frac{1}{2}    

   \Rightarrow \sin \frac{\pi}{6} = \frac{1}{2}  . 

  Hence y = \frac{\pi}{6}.

The required sum will be =     \frac{\pi}{3}+2(\frac{\pi}{6}) = \frac{2\pi}{3}.

Question:13 If \sin^{-1}x = y then

    (A)    0\leq y \leq \pi

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

    (C)     0 < y < \pi

    (D)     -\frac{\pi}{2} < y < \frac{\pi}{2}

Answer:

Given if  \sin^{-1}x = y  then,

As we know that the \sin^{-1} can take values between \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ].

Therefore, -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}.

Hence answer choice (B) is correct.

Question:14 \tan^{-1}(\sqrt3)-\sec^{-1}(-2)  is equal to

    (A)       \pi

    (B)   -\frac{\pi}{3}

    (C)       \frac{\pi}{3}

    (D)    \frac{2\pi}{3}

Answer:

Let us assume the values of \tan^{-1}(\sqrt3) be 'x'  and \sec^{-1}(-2)  be 'y'.

Then we have;

\tan^{-1}(\sqrt3) = x     or   \tan x = \sqrt 3    or  \tan \frac{\pi}{3} = \sqrt 3   or 

x = \frac{\pi}{3}

and  \sec^{-1}(-2) = y    or  \sec y = -2              

or   -\sec (\frac{\pi}{3}) =\sec ({\pi - \frac{\pi}{3}}) = \sec{\frac{2\pi}{3}}    

y = \frac{2\pi}{3}

also, the ranges of the principal values of \tan^{-1} and  \sec^{-1}   are (\frac{-\pi}{2},\frac{\pi}{2}). and 

[0,\pi] - \left \{ \frac{\pi}{2} \right \}   respectively.

\therefore we have then;

 \tan^{-1}(\sqrt3)-\sec^{-1}(-2) 

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

CBSE 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)                                        \left ( \because 3\sin \theta - 4\sin^3 \theta \right = \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 = \Thetathen 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).

Solutions of NCERT for class 12 maths chapter 2 Inverse Trigonometric Functions: Miscellaneous Exercise 

Question:1 Find the value of the following: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )

Answer:

If x \epsilon [0,\pi]  then  \cos^{-1}(\cos x) = x , which is principal value of \cos^{-1} x.

So, we have \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )

 where \:\frac{13\pi}{6} \notin \left [ 0, \pi \right ].

Hence\: we \:can\: write\: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) \:as

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

=\cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right )

   \frac{\pi}{6}\ \epsilon \left [ 0, \pi \right ]

Therefore we have,

\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) = \cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right ) = \frac{\pi}{6}.

Question:2 Find the value of the following: \tan^{-1}\left(\tan\frac{7\pi}{6} \right )

Answer:

We have given \tan^{-1}\left(\tan\frac{7\pi}{6} \right );

so, as we know  \tan^{-1}\left(\tan x \right ) =x \:\:if\:\:x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )

So, here we have \frac{7\pi}{6} \notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ).

Therefore we can write \tan^{-1}\left(\tan\frac{7\pi}{6} \right ) as:

=\tan^{-1}\left(\tan \left (2\pi - \frac{5\pi}{6} \right ) \right )           \left [ \because \tan(2\pi - x) = -\tan x \right ]

=\tan^{-1}\left[-\tan \left ( \frac{5\pi}{6} \right ) \right ]

=\tan^{-1}\left[\tan \left (\pi- \frac{5\pi}{6} \right ) \right ]

=\tan^{-1}\left[\tan \left (\frac{\pi}{6} \right ) \right ]\:\:where\:\:\frac{\pi}{6} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ) 

\therefore \tan^{-1} \left ( \tan \frac{7\pi}{6} \right ) = \tan^{-1} \left ( \tan \frac{\pi}{6} \right ) = \frac{\pi}{6}.

Question:3 Prove that 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}

Answer:

To prove: 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7};

L.H.S=2\sin^{-1}\frac{3}{5}

Assume that  \sin^{-1}\frac{3}{5} = x 

then we have \sin x = \frac{3}{5}.

or \cos x = \sqrt{1-\left (\frac{3}{5} \right )^2} = \frac{4}{5}

Therefore we have

 \tan x = \frac{3}{4}\:\:or\:\:x = \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}

Now,

We can write L.H.S as

 2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4}

 =\tan^{-1} \left [\frac{2\times\frac{3}{4}}{1- \left ( \frac{3}{4} \right )^2} \right ]    as we know    \left [2\tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \right ]

 =\tan^{-1} \left [\frac{\frac{3}{2}}{\left ( \frac{16-9}{16} \right )} \right ] = \tan^{-1}\left ( \frac{3}{2}\times \frac{16}{7} \right )

 =\tan^{-1} \frac{24}{7}=R.H.S  

L.H.S = R.H.S

Question:4 Prove that \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} =\tan^{-1}\frac{77}{36}

Answer

Taking \sin ^{-1} \frac{8}{17} = x  

then,

\sin x = \frac{8}{17} \Rightarrow \cos x = \sqrt{1- \left ( \frac{8}{17} \right )^2} = \sqrt {\frac{225}{289}} = \frac{15}{17}.

Therefore we have-

\tan^{-1} x = \frac{8}{15} \Rightarrow x = \tan^{-1} \frac{8}{15}    

 \therefore \sin ^{-1} \frac{8}{17} = \tan ^{-1} \frac{8}{15}        .............(1).

Now, let\:\sin ^{-1} \frac{3}{5} = y,

Then,

\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}                    .............(2).

So, we have now,

L.H.S.

\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5}

using equations (1) and (2) we get,

 =\tan ^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}

 =\tan^{-1} \frac{\frac{8}{15}+ \frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}                   [\because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1 - xy}] 

=\tan^{-1} (\frac{32+45}{60-24})

 =\tan^{-1} (\frac{77}{36}) 

=  R.H.S.

Question:5 Prove that \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}

Answer:

Take \cos^{-1}\frac{4}{5} = x  and \cos^{-1}\frac{12}{13} = y    and \cos^{-1}\frac{33}{65} = z

then we have,

\cos x = \frac{4}{5}

\sin x = \sqrt {1- \left ( \frac {4}{5} \right )^2} = \frac {3}{5}

Then we can write it as:

\tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}    or    x= \tan^{-1} \frac{3}{4}

\therefore \cos ^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}                                   ...............(1)

Now, \cos^{-1}\frac{12}{13} = y  

\cos y = \frac{12}{13} \Rightarrow \sin y =\frac{5}{13}

\therefore \tan y = \frac{5}{12} \Rightarrow y = \tan^{-1} \frac{5}{12}

So,  \cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}                        ...................(2)

Also we have similarly;

 \cos^{-1}\frac{33}{65} = z

Then,

\cos^{-1}\frac{33}{65} = \tan^{-1} \frac{56}{33}             ...........................(3)

Now, we have

L.H.S

\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13}    so, using (1) and (2) we get,

 =\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12}

 =\tan^{-1}\left ( \frac{\frac{3}{4}+ \frac{5}{12}}{1-\left ( \frac{3}{4}\times \frac{5}{12} \right )} \right )             \because \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]

=\tan^{-1}\left ( \frac{36+20}{48-15} \right )

 =\tan^{-1}\left ( \frac{56}{33} \right )    or we can write it as;

 =\cos^{-1}\frac{33}{65} 

=  R.H.S.

Hence proved.

 

Question:6 Prove that \cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}

Answer:

Converting all terms in tan form;

Let \cos^{-1}\frac{12}{13} = x ,  \sin^{-1}\frac{3}{5} = y  and \sin^{-1}\frac{56}{65} = z.

now, converting all the terms:

\cos^{-1}\frac{12}{13} = x  or   \cos x = \frac{12}{13}

We can write it in tan form as:

\cos x = \frac{12}{13} \Rightarrow \sin x = \frac{5}{13}.

\therefore \tan x = \frac{5}{12} \Rightarrow x = \tan^{-1} \frac{5}{12}

or \cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}           ................(1)

\sin^{-1}\frac{3}{5} = y     or   \sin y = \frac{3}{5}

We can write it in tan form as:

\sin y = \frac{3}{5} \Rightarrow\cos y = \frac{4}{5}

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

or \sin^{-1}\frac{3}{5} = \tan^{-1} \frac{3}{4}         ......................(2)

Similarly, for \sin^{-1}\frac{56}{65} = z;

we have \sin^{-1}\frac{56}{65} = \tan^{-1} \frac{56}{33}     .............(3)

Using (1) and (2) we have L.H.S

\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5}

= \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{3}{4}

On applying \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}

We have,

 =\tan^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-(\frac{5}{12}.\frac{3}{4})}

=\tan^{-1} (\frac{20+36}{48-15})

=\tan^{-1} (\frac{56}{33})

=\sin^{-1} (\frac{56}{65})                               ...........[Using (3)]

=R.H.S.

Hence proved.

Question:7 Prove that \tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}

Answer:

Taking R.H.S;

We have \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}

Converting sin and cos terms in tan forms:

Let \sin^{-1}\frac{5}{13} = x   and   \cos^{-1}\frac{3}{5} = y

now, we have \sin^{-1}\frac{5}{13} = x   or    \sin x = \frac{5}{13}

\sin x = \frac{5}{13} \:or\: \cos x =\frac{12}{13}\:or\:\tan x = \frac{5}{12}  

\tan x = \frac{5}{12} \Rightarrow x =\tan^{-1} \frac{5}{12}

\therefore \sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}                 ............(1)

Now, \cos^{-1}\frac{3}{5} = y\Rightarrow \cos y = \frac{3}{5}        

\cos y = \frac{3}{5} \:or\: \sin y = \frac{4}{5}\:or\:\tan y = \frac{4}{3}  

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

\therefore \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{4}{5}                 ................(2)

Now, Using (1) and (2) we get,

R.H.S.

\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}

=\tan^{-1}\left ( \frac{\frac{5}{12}+\frac{4}{3}}{1- \frac{5}{12}\times \frac{4}{3}} \right )    as we know \left [ \tan^{-1} x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy} \right ]

so,

= \tan^{-1} \frac{63}{16}

equal to L.H.S

Hence proved.

Question:8 Prove that \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} +\tan^{-1}\frac{1}{3} +\tan^{-1}\frac{1}{8} = \frac{\pi}{4}

Answer:

Applying the formlua:

\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}    on two parts.

we will have,

=\tan^{-1}\left (\frac{\frac{1}{5}+ \frac{1}{7}}{1- \frac{1}{5}\times \frac{1}{7}} \right ) + \tan^{-1}\left (\frac{\frac{1}{3}+ \frac{1}{8}}{1- \frac{1}{3}\times \frac{1}{8}} \right )

= \tan^{-1} \left ( \frac{7+5}{35-1} \right ) + \tan^{-1} \left ( \frac{8+3}{24-1} \right )

= \tan^{-1} \left ( \frac{12}{34} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )

= \tan^{-1} \left ( \frac{6}{17} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )

 = \tan^{-1}\left [ \frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\times\frac{11}{23}} \right ]

 = \tan^{-1}\left [ \frac{325}{325} \right ] = \tan^{-1} 1

=\frac{\pi}{4}

Hence it s equal to R.H.S

Proved.

Question:9 Prove that \tan^{-1} \sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x},\;\;x\in [0,1]

Answer:

By observing the square root we will first put

x= \tan^2 \theta.

Then,

we have \tan^{-1} \sqrt{\tan^2 \theta} = \frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta}

or, R.H.S.

\frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1}{2}\cos^{-1}(cos2 \theta)

= \frac{1}{2}\times 2\theta = \theta.

L.H.S. \tan^{-1} \sqrt{\tan^2 \theta} = \tan^{-1}(\tan \theta) = \theta

hence L.H.S. = R.H.S proved.

Question:10 Prove that \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \frac{x}{2},\;\;x\in\left(0,\frac{\pi}{4} \right )

Answer:

Given that \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )

By observing we can rationalize the fraction

        \left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )

We get then,

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

= \left(\frac{{1+\sin x} +{1 - \sin x} + 2\sqrt{(1+\sin x)(1-\sin x)} }{{1+\sin x} - {1 + \sin x}} \right )

= \frac{2(1+\sqrt{1-\sin^2 x})}{2\sin x} = \frac{1+\cos x}{\sin x} = \frac{2\cos^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}

= \cot \frac{x}{2}

Therefore we can write it as;

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

As L.H.S. = R.H.S.

Hence proved.

Question:11 Prove that \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right ) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x,\;\;-\frac{1}{\sqrt2}\leq x\leq 1

            [Hint: Put x = \cos 2\theta]

Answer:

By using the Hint we will put x = \cos 2\theta;

we get then,

=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right )

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

=\tan^{-1}\left(\frac{\sqrt2{\cos \theta} - \sqrt2{\sin\theta}}{\sqrt2{\cos \theta} + \sqrt2{\sin\theta}} \right )

=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right )   dividing numerator and denominator by \cos \theta,

we get,

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

= \tan^{-1} 1 - \tan^{-1} (\tan \theta)      using the formula  \left [ \tan^{-1}x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \right ]

= \frac{\pi}{4} - \theta = \frac{\pi}{4}- \frac{1}{2}\cos^{-1}x  

As L.H.S = R.H.S

Hence proved

Question:12 Prove that \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}

Answer:

We have to solve the given equation:

\frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3} 

Take \frac{9}{4} as common in L.H.S,

=\frac{9}{4}\left [ \frac{\pi}{2}- \sin^{-1}\frac{1}{3} \right ]

or =\frac{9}{4}\left [ \cos^{-1}\frac{1}{3} \right ]    from      \left [ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \right ]

Now, assume,

  \left [ \cos^{-1}\frac{1}{3} \right ] = y

Then,

\cos y = \frac{1}{3} \Rightarrow \sin y = \sqrt{1-(\frac{1}{3})^2} = \frac{2.\sqrt2}{3}

Therefore we have now,

y = \sin^{-1} \frac{2.\sqrt2}{3}

So we have L.H.S then = \frac{9}{4}\sin^{-1} \frac{2.\sqrt2}{3}

That is equal to R.H.S.

Hence proved.

Question:13 Solve the following equations: 2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)

Answer:

Given equation 2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x);

Using the formula:

\left [ 2\tan^{-1}z = \tan^{-1} \frac{2z}{1-z^2} \right ]

We can write

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

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

So, we can equate;

=\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \left [2cosec x \right ]

=\left [ \frac{2\cos x}{\sin^2 x }\right ] = \left [ \frac{2}{sinx } \right ]

that implies that \cos x = \sin x.

or    \tan x =1      or    x = \frac{\pi}{4}

Hence we have solution x = \frac{\pi}{4}.

Question:14 Solve the following equations: \tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x,\;(x>0)

Answer:

Given equation is

\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x:

L.H.S can be written as;

\tan^{-1} \frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x 

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

So, we have \tan^{-1}1 - \tan^{-1}x = \frac{1}{2} \tan^{-1}x

\Rightarrow \tan^{-1}1= \frac{3}{2} \tan^{-1}x

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

\Rightarrow \tan^{-1}x = \frac{\pi}{6}

 \Rightarrow x= \tan \frac{\pi}{6} = \frac{1}{\sqrt3}

Hence the value of x= \frac{1}{\sqrt3}.

Question:15 \sin(\tan^{-1}x),\;|x|<1  is equal to 

        (A)    \frac{x}{\sqrt{1-x^2}}

        (B)    \frac{1}{\sqrt{1-x^2}}

        (C)    \frac{1}{\sqrt{1+x^2}}

        (D)    \frac{x}{\sqrt{1+x^2}}

Answer:

Let \tan^{-1}x = y then we have;

\tan y = x   or

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

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

Hence the correct answer is D.

Question:16 \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2} then x is equal to

        (A)    0,\frac{1}{2}

        (B)    1,\frac{1}{2}

        (C)    0

        (D)    \frac{1}{2}        

Answer:

Given the equation: \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}

we can migrate the \sin^{-1}(1-x) term to the R.H.S.

then we have;

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

or - 2\sin^{-1}x =\cos^{-1}(1-x)                               ............................(1)

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

Take \sin^{-1}x = \Theta  \Rightarrow \sin \Theta = x    or    \cos \Theta = \sqrt{1-x^2}.

So, we conclude that;

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

Therefore we can put the value of \sin^{-1}x in equation (1)  we get,

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

Putting x= sin y, in the above equation; we have then,

\Rightarrow - 2\cos^{-1}\left ( \sqrt{1-(\sin y)^2} \right ) =\cos^{-1}(1-\sin y )

\Rightarrow - 2\cos^{-1}\left ( \sqrt{\cos^2 y} \right ) =\cos^{-1}(1-\sin y )

\Rightarrow - 2\cos^{-1}\left ( \cos y \right ) =\cos^{-1}(1-\sin y )

\Rightarrow \cos(-2y) = 1-\sin y

\Rightarrow - 2y=\cos^{-1}(1-\sin y )

 \Rightarrow 1- 2\sin^2 y = 1-\sin y

\Rightarrow 2\sin^2 y - \sin y = 0

\Rightarrow \sin y(2 \sin y -1) = 0

So, we have the solution;

\sin y = 0\ or\ \frac{1}{2}    Therefore we have x = 0\ or\ x= \frac{1}{2}.

When we have x= \frac{1}{2}, we can see that :

L.H.S. = \sin ^{-1}\left ( 1 - \frac{1}{2} \right ) - 2\sin^{-1}\frac{1}{2} = - \sin^{-1}\frac{1}{2} = -\frac{\pi}{6}

So, it is not equal to the R.H.S. -\frac{\pi}{6} \neq \frac{\pi}{2}

Thus we have only one solution which is x = 0

Hence the correct answer is (C).

Question:17\tan^{-1}\left (\frac{x}{y} \right )-\tan^{-1}\frac{x-y}{x+y} is equal to 

        (A)    \frac{\pi}{2}

        (B)    \frac{\pi}{3}

        (C)    \frac{\pi}{4}

        (D)    \frac{3\pi}{4}

Answer:

Applying formula: \left [ \tan^{-1} x - \tan^{-1}y = \tan^{-1} \left ( \frac{x-y}{1+xy} \right ) \right ].

We get, 

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

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

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

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

Hence, the correct answer is C. 

NCERT solutions for class 12 maths chapter-wise

chapter 1

Solutions of NCERT for class 12 maths chapter 1 Relations and Functions

chapter 2

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

chapter 3

NCERT solutions for class 12 maths chapter 3 Matrices

chapter 4

CBSE NCERT solutions for class 12 maths chapter 4 Determinants

chapter 5

Solutions of NCERT for class 12 maths chapter 5 Continuity and Differentiability

chapter 6

CBSE NCERT solutions for class 12 maths chapter 6 Application of Derivatives

chapter 7

NCERT solutions for class 12 maths chapter 7 Integrals

chapter 8

Solutions of NCERT for class 12 maths chapter 8 Application of Integrals

chapter 9

CBSE NCERT solutions for class 12 maths chapter 9 Differential Equations

chapter 10

NCERT solutions for class 12 maths chapter 10 Vector Algebra

chapter 11

Solutions of NCERT for class 12 maths chapter 11 Three Dimensional Geometry

chapter 12

CBSE NCERT solutions for class 12 maths chapter 12 Linear Programming

chapter 13

NCERT solutions for class 12 maths chapter 13 Probability

NCERT solutions for class 12

Students who wish to perform well in the CBSE 12 board examination, solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions are very helpful but here are some tips to make command on Inverse Trigonometric Functions.

  • The inverse trigonometric function is inverse of the trigonometric function, so if you have a command on the trigonometric function then it will be easy for you to understand inverse trigonometric functions. 
  • Try to relate trigonometric functions formulas with inverse trigonometric functions formulas, so that memorizing the formulae becomes easier. 
  • Before starting to solve exercise, first solve the examples that are given in the NCERT class 12 maths textbook.
  • Also, try to solve every exercise including miscellaneous exercise, NCERT chapter examples, miscellaneous examples on your own, if you are finding difficulties, you can take the help of CBSE NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions.
  • If you have solved all NCERT then you can solve previous years paper CBSE board to get familiar with the pattern of board exam question paper

Happy learning!!!

 

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