NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

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NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Q: 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.

Q: 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.

Q: 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.

Q: 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.

Q: 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.

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.

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Inverse Trigonometric Functions Class 12 Questions And Answers
Inverse Trigonometric Functions Class 12 Questions And Answers PDF Free Download
Inverse Trigonometric Functions Class 12 NCERT Solutions - Important Formulae
Class 12 Inverse Trigonometric Functions NCERT Solutions (Intext Questions and Exercise)
Class 12 inverse trigonometric functions NCERT solutions - Topics
Class 12 Maths Question Answer - Chapter Wise
Class 12 Maths Question Answer - Subject Wise
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

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 Questions And Answers PDF Free Download

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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';

$\tan^{-1}(-1) =z$ ,

$\tan z = -1$

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

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

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

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

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) $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

(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) $-\frac{\pi}{3}$

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

(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)

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

(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) $\frac{1}{\sqrt{1-x^2}}$

(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) $1,\frac{1}{2}$

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

(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

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Mathematics Solutions

NCERT Exemplar Class 12 Biology Solutions

NCERT Exemplar Class 12 Physics Solutions

Class 12 Maths Question Answer - Chapter Wise

Chapter 1	NCERT Solutions 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	NCERT solutions for class 12 maths chapter 4 Determinants
Chapter 5	NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability
Chapter 6	NCERT solutions for class 12 maths chapter 6 Application of Derivatives
Chapter 7	NCERT solutions for class 12 maths chapter 7 Integrals
Chapter 8	NCERT solutions for class 12 maths chapter 8 Application of Integrals
Chapter 9	NCERT solutions for class 12 maths chapter 9 Differential Equations
Chapter 10	NCERT solutions for class 12 maths chapter 10 Vector Algebra
Chapter 11	NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry
Chapter 12	NCERT solutions for class 12 maths chapter 12 Linear Programming
Chapter 13	NCERT solutions for class 12 maths chapter 13 Probability

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.

Also Read

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

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NCERT Books for Class 12 Hindi 2024 - Download chapter wise PDFs here

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NCERT Books for Class 12 - Download All Subjects PDFs Here

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NCERT Syllabus for Class 12 2024 (All Subjects) - Download PDF Here

NCERT Syllabus for Class 6 to 12 (2023-24) - All Subjects PDF Download

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Biology - Check CBSE 12 Biology NCERT Chapterwise Solution

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NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

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Hope this information helps you.

Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

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

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer

Muskan Dhalwal 17 August,2022

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Popular CBSE Class 12th Questions

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

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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 3.8×10⁷J 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

Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

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Bio Medical Engineer

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Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

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GIS Expert

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Budget Analyst

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Underwriter

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Azure Administrator

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