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

Last updated date: 05th Aug 2024
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## NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 - Free PDF Download

Exercise 2.2 in this chapter focuses on the properties, principal values, and applications of inverse trigonometric functions. This exercise is designed to help students deepen their understanding of these functions, which are essential in various fields such as calculus, physics, and engineering. NCERT for Exercise 2.2 Class 12 Maths Solutions Free PDF download and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Maths Chapter 2 Inverse Trigonometric Functions Class 12 Ex 2.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 - Free PDF Download
2. Glance on NCERT Solutions Maths Chapter 2 Ex 2.2 Class 12 | Vedantu
3. Topics Covered in the NCERT Solutions for Class 12 Maths Chapter 2
4. Access NCERT Solutions for Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions
5. Access other Exercises of Class 12 Maths Chapter 2: Inverse Trigonometric Functions
6. CBSE Class 12 Maths Chapter 2 Other Study Materials
7. Chapter wise NCERT Solutions for Class 12 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 2 Ex 2.2 Class 12 | Vedantu

• Inverse trigonometric functions, also called inverse sine, cosine, tangent, etc., are the inverses of the basic trigonometric functions (sine, cosine, tangent, etc.).

• Trigonometric functions tell you what the ratio of sides is in a right triangle given an angle. Inverse trig functions do the opposite - tells you the angle measure when given the trigonometric ratio.

• Common inverse functions include sin⁻¹(x), cos⁻¹(x), and tan⁻¹(x).

• The principal value of an inverse trigonometric function is the specific value within a defined range.

• For $\sin^{-1}(x)$ , the principal value range is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

• For $\cos^{-1}(x)$ , the principal value range is [0, \pi].

• For $\tan^{-1}(x)$ , the principal value range is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

• Exercise 2.2 Maths Class 12 NCERT Solutions has overall 15 fully solved Questions

• Class 12 Maths Chapter 2 Exercise 2.2 likely involved applying these formulas to various problems.

## Topics Covered in the NCERT Solutions for Class 12 Maths Chapter 2

1. Properties  of Inverse Trigonometric Functions

• 1. sin⁻¹(sin(x)) = x for x in [−π/2, π/2]

• 2. cos⁻¹(cos(x)) = x for x in [0, π]

• 3. tan⁻¹(tan(x)) = x for x in [−π/2, π/2]

• 4. sin(sin⁻¹(x)) = x for x in [-1, 1]

• 5. cos(cos⁻¹(x)) = x for x in [-1, 1]

• 6. tan(tan⁻¹(x)) = x for all x

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

Refer to page 10 for Exercise 2.2 in the PDF.

1. Prove $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right),x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]$

Ans: To Prove $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$, where $x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]$

Let $x = \sin \theta .$ Then, ${\sin ^{ - 1}}x = \theta$.

We have given that,

R.H.S ${\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) = {\sin ^{ - 1}}\left( {3\sin \theta - 4{{\sin }^3}\theta } \right)$

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

$= 3\theta$

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

Hence proved.

2. Prove $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\dfrac{1}{2},1} \right]$

Ans: To Prove $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\dfrac{1}{2},1} \right]$

Let $x = \cos \theta$. Then, ${\cos ^{ - 1}}x = \theta$

We have given that,

R.H S =${\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$

$= {\cos ^{ - 1}}\left( {4{{\cos }^3}\theta - 3\cos \theta } \right)$

$= {\cos ^{ - 1}}(\cos 3\theta )$

$= 3\theta$

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

Hence proved.

3. Write the function in the simplest form: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}} - 1}}{x},x \ne 0$

Ans: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}$

By putting $x = \tan \theta \Rightarrow \theta = {\tan ^{ - 1}}x$

$\therefore {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}} - 1}}{x} = {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}$

$= {\tan ^{ - 1}}\left( {\dfrac{{\sec \theta - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{2{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}} \right)$

$= {\tan ^{ - 1}}\left( {\tan \dfrac{\theta }{2}} \right) = \dfrac{\theta }{2} = \dfrac{1}{2}{\tan ^{ - 1}}x$

4. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi$

Ans: Given that,${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi$

${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right)$

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

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

$= {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right)$

$= \dfrac{x}{2}$

5. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right),0 < x < \pi$

Ans: We have given that, ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right)$

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

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

$= {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}(\tan x)\quad \left[ {\because \quad \dfrac{{ - y}}{{x - xy}} = {{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right]$

$= \dfrac{\pi }{4} - x$

6. Write the function in the simplest form: ${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }},|x| < a$

Ans: We have given that,${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

Let’s consider, $x = a\sin \theta \Rightarrow \dfrac{x}{a} = \sin \theta \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

$\therefore {\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

$= {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)$

$= {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\sin }^2}\theta } }}} \right)$

$= {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\cos \theta }}} \right)$

$= {\tan ^{ - 1}}(\tan \theta ) = \theta = {\sin ^{ - 1}}\dfrac{x}{a}$

7. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),a > 0;\dfrac{{ - a}}{{\sqrt 3 }} \leqslant x \leqslant \dfrac{a}{{\sqrt 3 }}$

Ans: Consider, ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

Let’s consider, $x = a\tan \theta \Rightarrow \dfrac{x}{a} = \tan \theta \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

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

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

$= {\tan ^{ - 1}}(\tan 3\theta )$

$= 3\theta$

$= 3{\tan ^{ - 1}}\dfrac{x}{a}$

8. Find the value of ${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\dfrac{1}{2}} \right)} \right]$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{1}{2} = x$

Then $\sin x = \dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)$

$\therefore {\sin ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{6}$

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

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

$= {\tan ^{ - 1}}\left[ {2\cos \dfrac{\pi }{3}} \right]$

$= {\tan ^{ - 1}}\left[ {2 \times \dfrac{1}{2}} \right]$

$= {\tan ^{ - 1}}1 = \dfrac{\pi }{4}$

9. Find the value of $\tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right],|x| < 1,y > 0$ and $xy < 1$

Ans: Let consider, $x = \tan \theta$.

Then, $\theta = {\tan ^{ - 1}}x$.

$\therefore {\sin ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} = {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$= {\sin ^{ - 1}}(\sin 2\theta )$

$= 2\theta$

$= 2{\tan ^{ - 1}}x$

Let’s assume, $y = \tan \theta .$ Then, $\theta = {\tan ^{ - 1}}y$.

$\therefore {\cos ^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)$

$= {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$= {\cos ^{ - 1}}(\cos 2\theta )$

$= 2\theta = 2{\tan ^{ - 1}}y$

$\therefore \tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)} \right]$

$= \tan \dfrac{1}{2}\left[ {2{{\tan }^{ - 1}}x + 2{{\tan }^{ - 1}}y} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

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

$= \dfrac{{x + y}}{{1 - xy}}$

10. Find the values of ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

As we know that ${\sin ^{ - 1}}(\sin x) = x$

If $x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, which is the principal value branch of ${\sin ^{ - 1}}x$.

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

Now, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$ can be written as:

${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

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

$= {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{3}} \right)$, where $\dfrac{\pi }{3} \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

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

11. Find the values of ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

Ans: Let’s Consider, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

As we know that ${\tan ^{ - 1}}(\tan x) = x$

If $x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$, which is the principal value branch of ${\tan ^{ - 1}}x$. Here, $\dfrac{{3\pi }}{4} \notin \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.

Now, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$ can be written as:

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

${\tan ^{ - 1}}\left( { - \tan \dfrac{\pi }{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right]$ where $- \dfrac{\pi }{4} \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

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

12. Find the values of $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{3}{5} = x$.

Then, $\sin x = \dfrac{3}{5}$

$\Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x} = \dfrac{4}{5}$

$\Rightarrow \sec x = \dfrac{5}{4}$

$\therefore \tan x = \sqrt {{{\sec }^2}x - 1} = \sqrt {\dfrac{{25}}{{16}} - 1} = \dfrac{3}{4}$

$\therefore x = {\tan ^{ - 1}}\dfrac{3}{4}$

$\therefore {\sin ^{ - 1}}\dfrac{3}{5} = {\tan ^{ - 1}}\dfrac{3}{4}\quad \ldots (i)$

Therefore, $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

$= \tan \left( {{{\tan }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{2}{3}} \right)\quad$ [Using (i) and (ii)]

$= \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{3}{4} \cdot \dfrac{2}{3}}}} \right)} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

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

$= \tan \left( {{{\tan }^{ - 1}}\dfrac{{17}}{6}} \right) = \dfrac{{17}}{6}$

13. Find the values of ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ is equal to

(A) $\dfrac{{7\pi }}{6}$

(B) $\dfrac{{5\pi }}{6}$

(C) $\dfrac{\pi }{3}$

(D) $\dfrac{\pi }{6}$

Ans: We know that ${\cos ^{ - 1}}(\cos x) = x$ if $x \in [0,\pi ]$, which is the principal value branch of ${\cos ^{ - 1}}x$. Here, $\dfrac{{7\pi }}{6} \notin [0,\pi ]$.

Now, ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ can be written as:

${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{ - 7\pi }}{6}} \right) = {\cos ^{ - 1}}\left[ {\cos \left( {2\pi - \dfrac{{7\pi }}{6}} \right)} \right]\quad [\because$

$+ x) = \cos x]$

$\therefore {\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{6}} \right) = \dfrac{{5\pi }}{6}$

The correct answer is ${\text{B}}$.

14. Find the values of $\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right)$ is equal to

(A) $\dfrac{1}{2}$

(B) $\dfrac{1}{3}$

(C) $\dfrac{1}{4}$

(D) 1

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = x$.

Then, $\sin x = \dfrac{{ - 1}}{2} = - \sin \dfrac{\pi }{6} = \sin \left( {\dfrac{{ - \pi }}{6}} \right)$.

As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$. ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{\pi }{6}$

$\therefore \sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right) = \sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{{3\pi }}{6}} \right) = \sin \left( {\dfrac{\pi }{2}} \right) = 1$

The correct answer is ${\text{D}}$.

15. $\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$ is equal to

(A) $\pi$

(B) $-\dfrac{\pi}{2}$

(C) 0

(D) $2 \sqrt{3}$

Ans:

Suppose that,

$A=\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$

As we know that,

$\cot ^{-1}(-x)=\pi-\cot ^{-1}(x)$

So,

$\cot ^{-1}(-\sqrt{3})=\pi-\cot ^{-1}(\sqrt{3})$

Put the value in the equation (i),

$A=\tan ^{-1}(\sqrt{3})-\pi+\cot ^{-1}(\sqrt{3})$

As we know that,

$\cot ^{-1} x=\dfrac{\pi}{2}-\tan ^{-1} x$

So,

$\cot ^{-1}(\sqrt{3})=\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3})$

Put the value in the equation (ii),

$A=\tan ^{-1}(\sqrt{3})-\pi+\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3})$

$A=-\dfrac{\pi}{2}$

Hence, the correct option is B.

## Conclusion

NCERT Solutions for Maths Chapter 2 Inverse Trigonometric Functions, Ex2.2 class 12 by Vedantu provides clear explanations and step-by-step solutions for understanding the properties and applications of inverse trigonometric functions. This exercise emphasizes the importance of knowing the principal value ranges for inverse functions, as well as solving equations involving these functions. Students should focus on mastering these ranges and practicing how to simplify and solve related equations accurately. Previous exam papers typically include 2-3 questions from this topic, highlighting its significance in understanding the broader subject of trigonometry. Vedantu’s solutions are designed to build a strong foundation, ensuring students can handle these concepts confidently in exams.

## Access other Exercises of Class 12 Maths Chapter 2: Inverse Trigonometric Functions

 Exercises Number of Questions Exercise 2.1 14 Questions and Solutions

## Chapter wise NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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

1. For the tests, which examples and questions from class 12th maths exercise 2.2 are the most crucial?

Exercise 2.2 in math class 12 comprises 21 questions and 6 examples (Examples 3, 4, 5, 6, 7, 8). It is imperative to complete all of the exercise's examples and questions. But because these examples and questions have already been asked in the tests numerous times, examples 3, 5, and 6 and questions 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15, 18, 19, and 21 of exercise 2.2 of class 12th math are most crucial.

2. How many days are required to finish Math Exercise 2.2 for grade 12?

Students will need two to three days to do exercise 1.2 of class 12 math if they can devote two hours every day to it. This time is a rough estimate. Because no two students work at the same rate or with the same efficiency, this time may vary.

3. How can you simplify NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2?

It will be simpler for the pupils to learn if they comprehend it and practise it. They will also receive full practise with the questions and the procedure from Vedantu ncert solutions for class 12th exercise 1.2.   Students will benefit greatly from asking the same questions over.

4. Is exercise 2.2 from Maths class 12 difficult for the test?

Exercise 2.2 in math class 12 is neither simple nor difficult. Because some of the examples and questions in this exercise are simple and some are difficult, it falls somewhere in the middle of easy and tough. The amount of difficulty for any topic or question does, however, vary from child to child. Therefore, whether or not exercise 2.2 of math class 12 is difficult depends on the students. Some youngsters find it challenging, some find it simple, and yet others find it somewhere in the centre.

5. What is exercise 2.2 based upon?

In class 11 trigonometry, questions based on sin 2x, sin 3x, cos 3x,   etc. are found in exercise 2.2. It also has questions that use square roots to simplify inverse trigonometric functions. The answers to the trigonometric function-related questions 14 and 15 are listed below. Once the function has been solved, we should check the outcome to see if it satisfies the equation before deciding on the final response.