RD Sharma Class 12 Exercise 3.1 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 3.1 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 11:30 AM IST

RD Sharma is considered the best book for students studying in CBSE schools. It is highly informative, has many examples, and is easy to understand. Many teachers prefer this material for their lectures, and students refer to them for exam preparation. This is the most fundamental material that CBSE students should have for maths.

RD Sharma Class 12 Solutions Inverse Trigonometric Functions Exercise 3.1 has 14 questions, including subparts of Level 1 difficulty. However, these questions are fundamental concept-based and can be completed relatively quickly if you have gone through the theorems.

## RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Function Exercise 3.1 Question 1 (i)

$\frac{-\pi }{3}$
Hint:
$f\! \left ( x \right )=y$
$f\! ^{-1}\left ( y \right )=x$
Given:
$Find\: principal\: value\: o\! f \sin ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )$
Solution:
$Let,\sin ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )= y$
$\Rightarrow \sin y= \frac{-\sqrt{3}}{2}$
$\Rightarrow \sin y= -\sin \left ( \frac{\pi }{3} \right )= \sin \left ( \frac{-\pi }{3} \right )$
$Range\: o\! f\: principal \: value\: o\! f\: \sin ^{-1} is\left ( \frac{-\pi }{2},\frac{\pi }{2} \right )\: and \: \sin \left ( \frac{-\pi }{3} \right )= \frac{-\sqrt{3}}{2}$
$There\! f\! ore, principal \: value\: o\! f\: \sin ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )is\: \frac{-\pi }{3}$

$\frac{-\pi }{6}$
Hint:
$\cos \left ( 90^{\circ}+x \right )= -\sin x$
Given:
$Find\: principal\: value\: o\! f \sin^{-1}\left ( \cos \frac{2\pi }{3} \right )$
Solution:
$Let, \sin^{-1}\left ( \cos \frac{2\pi }{3} \right )= y$
$\Rightarrow \sin y= \left ( \cos \frac{2\pi }{3} \right )$
$\Rightarrow \sin y=\cos \frac{2\pi }{3}= \cos \left ( \frac{\pi }{2}+\frac{\pi }{6} \right )= -\sin\left ( \frac{\pi }{6} \right )$
$W\! e\, know\, principal\, value\, o\! f \sin^{-1} is\, \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and -\sin \sin \left ( \frac{\pi }{6} \right )= \cos \cos \left ( \frac{2\pi }{3} \right )$
$Therefore, principal\, value\, o\! f \, \sin^{-1}\left [ \cos \left ( \frac{2\pi }{3} \right ) \right ] is\, \frac{-\pi }{6}.$

Inverse Trigonometric Function Exercise 3.1 Question 1 (iii)

$\frac{\pi }{12}$
Hint:
$Separate\, and\,\, reorganize \,\, the\, values$
Given:
$Find \: principal\: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right )$
Solution:
$\sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right )$$= \sin^{-1}\left ( \frac{\sqrt{3}}{2\sqrt{2}} -\frac{1}{2\sqrt{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} -\frac{1}{2}\times \frac{1}{\sqrt{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2}\times \sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}}-\frac{1}{\sqrt{2}}\times \sqrt{1-\left ( \frac{\sqrt{3}}{2} \right )^{2}} \right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right )-\sin^{-1}\left ( \frac{1}{\sqrt{2}} \right )$
$= \frac{\pi }{3}-\frac{\pi }{4}$
$=\frac{\pi }{12}$
$There\! fore, principal \: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right ) is\: \frac{\pi }{2}.$

Inverse Trigonometric Function Exercise 3.1 Question 1 (iv)

$\frac{7\pi }{12}$
Hint:
$Separate\: and\: reorganize\: the\: values$
Given:
$Find\: principal\: value\: o\! f \sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right )$
Solution:
$\sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right )$$= \sin^{-1}\left ( \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \times \sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}}+\frac{1}{\sqrt{2}}\times \sqrt{1-\left ( \frac{\sqrt{3}}{2} \right )^{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right )+\sin^{-1}\left ( \frac{1}{\sqrt{2}} \right )$
$= \frac{\pi }{3}+\frac{\pi }{4}$
$= \frac{7\pi }{2}$
$Therefore, principal\: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right ) is = \frac{7\pi }{2}.$

Inverse Trigonometric Function Exercise 3.1 Question 1 (v)

$\frac{-\pi }{4}$
Hint:
$\cos x= - \cos \left ( 180^{\circ}-x \right )$
Given:
$Find\: principal\: value\: o\! f \sin^{-1}\left ( \cos \left ( \frac{3\pi }{4} \right ) \right )$
Solution:
$Let, \sin^{-1}\left ( \cos \left ( \frac{3\pi }{4} \right ) \right )= y$
$\Rightarrow\sin y = \cos \left ( \frac{3\pi }{4} \right )$
$\Rightarrow \sin y= \cos \frac{3\pi }{4}= \cos\left ( \pi -\frac{\pi }{4} \right )= - \cos\left ( \frac{\pi }{4} \right )$
$Principal\: value \: o\! f \cos^{-1}\: is\: \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]\: and\: -\cos \cos\left ( \frac{\pi }{4} \right ) = \cos \cos\left ( \frac{3\pi }{4} \right )$
$There\! f\! ore, principal\: value\: o\! f \sin^{-1}\left ( \cos \left ( \frac{3\pi }{4} \right ) \right ) is \frac{-\pi }{4}.$

Inverse Trigonometric Function Exercise 3.1 Question 1 (vi)

$\frac{\pi }{2}$
Hint:
$\tan \left ( \pi +x \right )= \tan x$

Given:
$Find\: principal\: value\: o\! f \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right )$

Solution:
$Let , y= \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right )$
$\sin y= \tan \left ( \frac{5\pi }{4} \right )$
$\sin y= \tan \left (\pi + \frac{\pi }{4} \right )$
$= \tan \left ( \frac{\pi }{4} \right )$
$=1$
$= \sin \left ( \frac{\pi }{2} \right )$
$Range\: Principal\: value\: o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \: \sin \left ( \frac{\pi }{2} \right )= \tan \left ( \frac{5\pi }{4} \right )$
$There\! f\! ore, principal\: value\: o\! f \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right ) is \: \frac{\pi }{2}.$

$\frac{-\pi }{3}$
Hint:
$2\sin^{-1}x= \sin^{-1}\left ( 2x\sqrt{1-x^{2}} \right )$
Given:
$Find\, principal \, value\, o\! f \sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}}$
Solution:
$\sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}}$
$= \sin^{-1}\frac{1}{2}-\sin^{-1}\left ( 2\times \frac{1}{\sqrt{2}}\sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}} \right )$
$= \sin^{-1}\frac{1}{2}-\sin^{-1}1$
$= \frac{\pi }{6}-\frac{\pi }{2}$
$= \frac{-\pi }{3}$
$There\! fore, principal\, value\, o\! f \sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}} \, is = \frac{-\pi }{3}.$

Inverse Trigonometric Function Exercise 3.1 Question 2 (ii)

$\frac{\pi }{6}$
Hint:
$\sin^{-1}x= y$
$\sin y= x$
Given:
$Find\: principal\, value \, o\! f \sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \}$
Solution:
$\sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \}$
$= \sin^{-1}\left [ \cos \left ( \frac{\pi }{3} \right ) \right ] \; \; \; \; \; \; \; \; \left [ \because \left ( \frac{\sqrt{3}}{2} \right ) = \frac{\pi }{3}\right ]$
$= \left ( \frac{1}{2} \right )\; \; \; \; \; \; \; \; \left [ \because \cos \frac{\pi }{3}= \frac{1}{2} \right ]$
$Let , \left ( \frac{1}{2} \right )= y$
$\sin \sin y =\frac{1}{2}$
$Range\, o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \, \sin \sin \left ( \frac{\pi }{6} \right ) =\frac{1}{2}$
$Therefore, principal \, value\, o\! f \, \sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \} is \, \frac{\pi }{6}.$

Inverse Trigonometric Function Exercise 3.1 Question 3 (i)

$\left [ -1,1 \right ]$
Hint:
$The\: domain\: o\! f \sin^{-1}x \: is \left [ -1,1 \right ]$
Given:
$\sin^{-1}x^{2}$
Explanation:
$Let \: f\! \left ( x \right )= \left ( x^{2} \right )$
$\! \! \! \! \! \! \! \! \! \! \! \! The\, domain \, o\! f \sin^{-1} is \left [ -1,1 \right ] which\, implies\, that \, the\, value\, o\! f \, x\, lies\, between -1 and\, 1$
$i.e; -1\leq x\leq 1$
$but\: here\: we\: have\: the\: f\! unction \: x^{2}$
$so,-1\leq x^{2}\leq 1\; \; \; \; \; \; \; [As\, x^{2}\: is\: always\: positive]$
$\Rightarrow 0\leq x^{2}\leq 1$
$\Rightarrow\left | x \right | \leq 1$
$\Rightarrow x\leq 1\: and x\geq -1\: \: \: \: \: \: \: \: \: \: \: \left [ \because a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right ) \right ]$
$Hence, the\: domain\: o\! f \sin^{-1}x^{2} \: is \left [ -1,1 \right ].$

$\left [ -1,1 \right ]$
Hint:
$The\: domain \: o\! f \sin^{-1}x \: is \left [ -1,1 \right ] and\: the\: domain\: o\! f \sin x\: is \: R .$
Given:
$f\! \left ( x \right )= \sin^{-1}x+\sin x$
Explanation:
$W\! hen \, we\, f\! ind\, the\, domain\, o\! f \, the \, f\! unction\, which \, is \, f\! urther\, the\, sum \, o\! f \, two\, di\! f\! f\! erent \, f\! unctions\, then\, we\, take\, the\, intersection \, o\! f \, thei\, r respective\, domains.$ $\left [-1,1 \right ]\cap R$
$= \left [ -1,1 \right ]\cap \left ( -\infty ,\infty \right )$
$= \left [ -1,1 \right ]$
Hence, the domain of $\sin^{-1}x+\sin x$ is $\left [ -1,1 \right ]$.

Inverse Trigonometric Function Exercise 3.1 Question 3 (iii)

$\left [ -\sqrt{2},1 \right ]\cup \left [ 1,\sqrt{2} \right ]$
Hint:
$The\: domain \: o\! f \sin^{-1}x \: is \left [ -1,1 \right ].$
Given:
$f\! \left ( x \right )= \sin^{-1}\sqrt{x^{2}-1}$
Explanation:
$-1\leq \sqrt{x^{2}-1}\leq 1$
$x^{2}-1\leq 1 \: [Squaring\: both \: sides]$
$x^{2}-1+1\leq 1+1$
$x^{2}\leq 2$
$-\sqrt{2}\leq x\leq \sqrt{2} \cdot \cdot \cdot \cdot (i)$
$Also, x^{2}-1\geq 0$
$x^{2}\geq 1$
$\Rightarrow x\leq -1 \: and \: x\geq 1 \cdot \cdot \cdot \cdot (ii)$
$From (i) and (ii), the\: value\: o\! f \: x \: lies\: between$
$1\leq x\leq\sqrt{2} \: and \: -\sqrt{2}\leq x\leq -1$
$The \: domain \: o\! f \sin^{-1}\sqrt{x^{2}-1} \: is \left [ -\sqrt{2,}-1 \right ]\cup \left [ 1,\sqrt{2} \right ]$

$\left [ \frac{-1}{2} ,\frac{1}2{}\right ]$
Hint:
$The \, domain\, o\! f \sin^{-1}x \, is \left [ -1,1 \right ].$
Given:
$f\! \left ( x \right )= \sin^{-1}x+\sin^{-1}2x$
Explanation:
$Now, we\, f\! ind\, the\, domain\, o\! f \sin^{-1}2x$
$-1\leq 2x\leq 1$
$\frac{-1}{2}\leq x\leq \frac{1}{2}$
$The\, domain\, o\! f \, f\! \left ( x \right ) is\, the\, inter\! section \, o\! f\, domains\, o\! f \sin^{-1}x\, and \, \sin^{-1}2x.$
$\left [ \frac{-1}2,\frac{1}2{}{} \right ]\cap \left [ -1,1 \right ]$
$= \left [ \frac{-1}2,\frac{1}2{}{} \right ]$
$4$
Given:
$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z+\sin^{-1}t= 2\pi \: then\, f\! ind\: x^{2}+y^{2}+z^{2}+t^{2}$
Hint:
$We \: know \: that\: the\: range\: o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$
$\sin^{-1}x\leq \frac{\pi }{2},\sin^{-1}y\leq \frac{\pi }{2},\sin^{-1}z\leq \frac{\pi }{2},\sin^{-1}t\leq \frac{\pi }{2}$
Solution:
$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z+\sin^{-1}t\leq 2\pi \: and\: it \: is\: given\: that \: L.H.S \: is\: 2\pi.$
$So\: it\: is\: possible\: only\: when$
$\sin^{-1}x= \frac{\pi }{2},\sin^{-1}y= \frac{\pi }{2},\sin^{-1}z= \frac{\pi }{2},\sin^{-1}t= \frac{\pi }{2}$
$x= 1,y= 1,z= 1,t= 1$
$\therefore x^{2}+y^{2}+z^{2}+t^{2}= 4$

Inverse Trigonometric Function Exercise 3.1 Question 5

$3$
Given:
$\left ( \sin^{-1}x \right )^{2}+\left (\sin^{-1}y \right )^{2}+\left ( \sin^{-1}z \right )^{2}= \frac{3}{4}\pi ^{2} \,\, then\, find \,\, x^{2}+y^{2}+z^{2}.$
Hint:
$We\, know\, that \, the\, range\, o\! f \, \sin^{-1} is \left [ \frac{-\pi }{2} ,\frac{\pi }{2}\right ]$
$\sin^{-1}x\leq \frac{\pi }{2},\sin^{-1}y \leq \frac{\pi}{2}, \sin^{-1}z\leq \frac{\pi}{2}$
$\left ( \sin^{-1}x \right )^{2}\leq \frac{\pi ^{2}}{4},\left (\sin^{-1}y \right )^{2}\leq \frac{\pi ^{2}}{4},\left ( \sin^{-1}z \right )^{2}\leq \frac{\pi ^{2}}{4}$
$\left ( \sin^{-1}x \right )^{2}+\left (\sin^{-1}y \right )^{2}+\left ( \sin^{-1}z \right )^{2}\leq \frac{3}{4}\pi ^{2} \, \, and \, it\, is\, given\, that\, L.H.S\, is \,\, \frac{3}{4}\pi ^{2}.$
$So\: it\: is\: possible\: only\: when$
$\sin^{-1}x= \frac{\pi }{2},\sin^{-1}y= \frac{\pi}{2}, \sin^{-1}z= \frac{\pi}{2}$
$x= 1,y= 1,z= 1$$\therefore x^{2}+y^{2}+z^{2}= 3$

The concepts used in this exercise are:

• Evaluating the Inverse of Sin and Cos functions

• Finding the domain of inverse functions

• Finding values of variables assigned to inverse equations

Students facing problems in understanding the RD Sharma textbook can use RD Sharma Class 12th Exercise 3.1 solutions to study for their exams.

RD Sharma Solutions can help you prepare for 12 board exams and other entrance exams involving Maths. Using RD Sharma Class 12 Solutions Chapter 3 Exercise 3.1 to practice questions has been shown to improve math skills.

Expert Mathematics faculties have solved RD Sharma Class 12th Solution Inverse Trigonometric Function Exercise 3.1. All the important questions that may be asked in competitive exams such as JEE are included in the solutions. Exercise Questions with Solutions from RD Sharma Class 12th Exercise 3.1 might help you revise the entire syllabus and get more marks.

Both trigonometry and Inverse trigonometric functions are vast chapters that contain hundreds of sums. However, if you divide the sums and practice daily, you can easily cover this chapter with maximum output.

As RD Sharma Class 12th Exercise 3.1 material complies with the CBSE syllabus, students can refer to it as a follow-up for their classes. Moreover, teachers obviously can't cover hundreds of sums in their lectures, so you can use this as a reference for the questions in RD Sharma books. Moreover, as this is expert-created material, it is ensured that the sums are accurate and informative.

You must discover the principal value of several trigonometric functions given in the questions in RD Sharma Class 12 Chapter 3 Exercise 3.1. Then, you may put these answers into practice and work on their problem-solving skills, which will help them gradually gain confidence. If you find other chapters difficult, you can use the links provided to get help with those.

Chapter-wise RD Sharma Class 12 Solutions

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1. Can I improve my grades through this material?

RD Sharma Class 12th Exercise 3.1 is exclusively intended to assist students with a comprehensive and complete chapter overview. This will definitely help you score good marks.

3. Where can I get solutions for RD Sharma Class 12th Exercise 3.1?

All you have to do is go to Career360's website and get the book in PDF format for free. The book's pdf version is always available and requires no subscription costs. You can download it and study from the comfort of your home.

4. Explain the Inverse of fundamental trigonometric values?

Inverse of Sin θ is Cosec θ, Inverse of Cos θ is Sec θ and Inverse of Tan θ is Cot θ. For more info on Inverse Trigonometry, you can refer to Class 12 RD Sharma Chapter 3 Exercise 3.1 Solution.

5. What are the uses of Inverse Trigonometric Functions?

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