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

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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

Kuldeep MauryaUpdated on 21 Jan 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.

Inverse Trigonometric Functions Excercise: 3.1

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

Inverse Trigonometric Function Exercise 3.1 Question 1 (i)

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

Inverse Trigonometric Function Exercise 3.1 Question 1 (ii)

Answer:
$\frac{-\pi }{6}$
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)

Answer:
$\frac{\pi }{12}$
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)

Answer:
$\frac{7\pi }{12}$
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)

Answer:
$\frac{-\pi }{4}$
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)

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

Inverse Trigonometric Function Exercise 3.1 Question 2 (i)

Answer:
$\frac{-\pi }{3}$
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)

Answer:
$\frac{\pi }{6}$
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)

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

Inverse Trigonometric Function Exercise 3.1 Question 3 (ii)

Answer:
$\left [ -1,1 \right ]$
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)

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

Inverse Trigonometric Function Exercise 3.1 Question 3 (iv

Answer:
$\left [ \frac{-1}{2} ,\frac{1}2{}\right ]$
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:
$Now, we\, f\! ind\, the\, domain\, o\! f \sin^{-1}2x$
$-1\leq 2x\leq 1$
$\frac{-1}{2}\leq x\leq \frac{1}{2}$
$The\, domain\, o\! f \, f\! \left ( x \right ) is\, the\, inter\! section \, o\! f\, domains\, o\! f \sin^{-1}x\, and \, \sin^{-1}2x.$
$\left [ \frac{-1}2,\frac{1}2{}{} \right ]\cap \left [ -1,1 \right ]$
$= \left [ \frac{-1}2,\frac{1}2{}{} \right ]$

Inverse Trigonometric Function Exercise 3.1 Question 4

Answer:
$4$
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

Answer:
$3$
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

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