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NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Edited By Ramraj Saini | Updated on Sep 13, 2023 08:44 PM IST | #CBSE Class 12th

Inverse Trigonometric Functions Class 12 Questions And Answers

Inverse Trigonometric Functions class 12 Maths solutions are provided here. These NCERT solutions are created by expert team at Careers360 considering the demand of class 12 students, CBSE latest syllabus and pattern 2023-23, and comprehensively covering step by step all the concepts. Therefore these are great resource for students to command the concepts and in-depth understanding of the concepts. These inverse trigonometric functions class 12 NCERT solutions is one of the most important chapters in the NCERT Class 12 Maths book. These class 12 Maths chapter 2 notes will be beneficial for board exams as well as for other entrance exams such as JEE.

Concepts of maths chapter 2 class 12 are useful in other concepts like differential calculus and Integral calculus therefore class 12 inverse trigonometry becomes very important for students as it is base. Inverse trigonometric functions of NCERT class 12 chapter 2 also include exercises that are important as they help in practicing the problems and in depth understanding of class 12 chapter 2 inverse trigonometric functions of ncert.

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Inverse Trigonometric Functions Class 12 NCERT Solutions - Important Formulae

>> Basic Concepts: Inverse trigonometric functions map real numbers back to angles.

The inverse of the sine function: sin-1(x) or arcsin(x) is defined on [-1, 1].

Properties of Inverse Trigonometric Functions:

Function

Domain

Range

y = sin-1(x)

[-1, 1]

[-π/2, π/2]

y = cos-1(x)

[-1, 1]

[0, π]

y = cosec-1(x)

R - (-1, 1)

[-π/2, π/2] - {0}

y = sec-1(x)

R - (-1, 1)

[0, π] - {π/2, π/2}

y = tan-1(x)

R

(-π/2, π/2)

y = cot-1(x)

R

(0, π)

Self-Adjusting Trigonometric Property:

sin(sin-1(x)) = x

sin-1(sin(x)) = x

cos(cos-1(x)) = x

cos-1(cos(x)) = x

tan(tan-1(x)) = x

tan-1(tan(x)) = x

sec(sec-1(x)) = x

sec-1(sec(x)) = x

cosec-1(cosec(x)) = x

cosec(cosec-1(x)) = x

cot-1(cot(x)) = x

cot(cot-1(x)) = x

Reciprocal Relations:

sin-1(1/x) = cosec-1(x), x ≥ 1 or x ≤ -1

cos-1(1/x) = sec-1(x), x ≥ 1 or x ≤ -1

tan-1(1/x) = cot-1(x), x > 0

Even and Odd Functions:

sin-1(-x) = -sin-1(x), x ∈ [-1, 1]

tan-1(-x) = -tan-1(x), x ∈ R

cosec-1(-x) = -cosec-1(x), |x| ≥ 1

cos-1(-x) = π - cos-1(x), x ∈ [-1, 1]

sec-1(-x) = π - sec-1(x), |x| ≥ 1

cot-1(-x) = π - cot-1(x), x ∈ R

Complementary Relations:

sin-1(x) + cos-1(x) = π/2

tan-1(x) + cot-1(x) = π/2

cosec-1(x) + sec-1(x) = π/2

Sum and Difference Formulas:

tan-1(x) + tan-1(y) = tan-1((x+y)/(1-xy))

tan-1(x) - tan-1(y) = tan-1((x-y)/(1+xy))

sin-1(x) + sin-1(y) = sin-1[x√(1-y2)+y√(1-x2)]

sin-1(x) - sin-1(y) = sin-1[x√(1-y2)-y√(1-x2)]

cos-1(x) + cos-1(y) = cos-1[xy-√(1-x2)√(1-y2)]

cos-1(x) - cos-1(y) = cos-1[xy+√(1-x2)√(1-y2)]

cot-1(x) + cot-1(y) = cot-1((xy-1)/(x+y))

cot-1(x) - cot-1(y) = cot-1((xy+1)/(y-x))

Double Angle Formula:

2 tan-1(x) = sin-1(2x/(1+x2))

2 tan-1(x) = cos-1((1-x2)/(1+x2))

2 tan-1(x) = tan-1(2x/(1-x2))

2 sin-1(x) = sin-1(2x√(1+x2))

2 cos-1(x) = sin-1(2x√(1-x2))

Conversion Properties:

sin-1(x) = cos-1(√(1-x2)) = tan-1(x/√(1-x2)) = cot-1(√(1-x2)/x)

cos-1(x) = sin-1(√(1-x2)) = tan-1(√(1-x2)/x) = cot-1(x/√(1-x2))

tan-1(x) = sin-1(x/√(1-x2)) = cos-1(x/√(1+x2)) = sec-1(√(1+x2)) = cosec-1(√(1+x2)/x)

Free download Inverse Trigonometric Functions Class 12 NCERT solutions for CBSE Exam.

Class 12 Inverse Trigonometric Functions NCERT Solutions (Intext Questions and Exercise)

Inverse Trigonometric Functions Class 12 Solutions: 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}

Inverse Trigonometric Functions Class 12 Solutions: 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).


NCERT solutions 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.

If you are looking for inverse trigonometric functions class 12 exercises then these are listed below.

Inverse Trigonometric Functions Class 12 Exercise 2.1

Inverse Trigonometric Functions Class 12 Exercise 2.2

Inverse Trigonometric Functions Class 12 Miscellaneous Exercise

NCERT Solutions for Class 12 Maths Chapter 2 Introduction

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 exercises given in inverse trigonometric functions class 12. You may find some difficulties in solving the problems, so you can take the help of inverse trigonometry class 12 solutions maths chapter 2 notes.

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 ch 2 question answer function 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.

Class 12 inverse trigonometric functions NCERT solutions - Topics

The important topics covered in the class 12 maths ch 2 question answer are

  • Basic concepts

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This chapter of class 12 NCERT includes concepts of domain and range of trigonometric function in which their inverse exists. the chapter comprehensively discusses graphical representation, the elementary property of inverse trigonometry function.

ch 2 maths class 12 discusses the basic concepts of the inverse of the sine function, cosine function, secant function, cosecant function, tangent function, and cotangent function. Also, we will be familiar with their range and domain, concepts related to one-one, and onto function. Problems related to these concepts are in detail in NCERT solutions for class 12 maths chapter 2.

  • Properties of inverse trigonometry function

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Maths class 12 chapter 2 includes important property of inverse trigonometry function that might be valid in principal value branch and corresponding inverse trigonometric function. To get a good hold on these concepts you can solve the NCERT solution for class 12 maths chapter 2.

Topics mentioned in class 12 ncert are very important to obtain decent marks in board exams and students are recommended to proceed through all the concepts cited in the topics. Questions from all the above topics are covered in the NCERT solutions for class 12 maths chapter 2.

NCERT Exemplar Class 12 Solutions

Class 12 Maths Question Answer - Chapter Wise

More about NCERT Solutions for Class 12 Maths Chapter 2

  • In class 11 maths you have already learnt about trigonometric functions. It won't take much effort to command inverse trigonometric functions if you have good knowledge of trigonometric functions.

  • For this, 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 NCERT Class 12 Maths solutions chapter 2.

  • NCERT Solutions for Class 12 Maths Chapter 2 PDF Download will also be available soon to help you with the questions offline.

Inverse Trigonometric Functions Class 12 Solutions - Topics

2.1 Introduction

2.2 Basic Concepts

2.3 Properties of Inverse Trigonometric Functions

Class 12 Maths Question Answer - Subject Wise

NCERT Books and NCERT Syllabus

Tips to use class 12 maths chapter 2 question answer

  • If you wish to perform well in the CBSE 12 board examination, NCERT Class 12 Maths solutions 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, use class 12 maths chapter 2 question answer pdf to learn easily.

  • Try to relate trigonometric functions formulas with inverse trigonometric functions formulas, so that memorizing the formulae becomes easier.

  • Before starting to solve an exercise, first solve the examples that are given in the NCERT class 12 maths textbook.

  • Also, try to solve every exercise including inverse trigonometric functions class 12 miscellaneous, and examples, miscellaneous examples on your own, if you are finding difficulties, you can take the help of Class 12 Maths Chapter 2 NCERT solutions.

  • If you have solved all NCERT then you can solve previous years paper CBSE board to get familiar with the pattern of the board exam question paper.

Also, check NCERT solutions for class 12 for other chapter and subjects.

Frequently Asked Question (FAQs)

1. What are the important topics in the chapter on inverse trigonometric functions?

Some basic concepts of inverse trigonometry and properties of inverse trigonometric functions are important topics in this chapter. these concepts are very important as they are used in solving or simplifying differential calculus, integral calculus, and other topics like differential equations therefore having a command on inverse trigonometry concepts becomes very important. students can refer to class 12 maths chapter 2 question answer for getting a good hold on the concepts.

2. Do inverse trigonometric functions class 12 solutions aid in preparing for the board exams?

NCERT class 12 maths ch 2 solutions provide students with a strong foundation in basic Maths topics. These exercise-wise class 12 maths ch 2 ncert solutions are crafted by experienced and knowledgeable subject experts at Carresr360. By using these resources, students can improve their focus and achieve better scores on their board exams. The ultimate goal is to enhance students' confidence and ability to efficiently solve difficult problems in a timely manner.

3. What is the weightage of the chapter inverse trigonometry for the CBSE board exam?

Relation and function and inverse trigonometry combined have 10 % weightage in the CBSE class 12 board examination. According to priority students can give time to chapters but if you want full marks then it demand practice and in depth understanding of concepts therefor ncert textbook are suggested to students.  

4. Which are the most difficult chapters of NCERT Class 12 Maths syllabus?

Students consider Integration and it's application as the most difficult unit in CBSE class 12 maths. With regular practice, you will get conceptual clarity and will be able to have a strong grip on Integration. if you want then inverse trigonometric functions solutions are recommended to students for getting command and in depth understanding of concepts.

5. Which is the best book for CBSE class 12 Maths?

NCERT textbook is the best book for CBSE class 12 maths. Most of the questions in the CBSE class 12 board exam are directly asked from the NCERT textbook. All you need to do is practice of the questions given in the NCERT textbook that is best way to get command on concepts. if you have any difficulty then you can read solutions of ncert exercise also help you in getting hold on the chapter.

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Have a question related to CBSE Class 12th ?

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

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

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hello,

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

I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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