RD Sharma Class 12 Exercise 3.12 Inverse Trigonometric Functions Solutions Maths

RD Sharma Class 12 Exercise 3.12 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Updated on Jan 21, 2022 11:37 AM IST

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

Inverse Trigonometric Functions Excercise: 3.12

Inverse Trigonometric Function Exercise 3.12 Question 1 .

Answer:$\frac{33}{65}$
Given:
$\cos \left ( \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13} \right )$
Hint: We will use the formula
$\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}} \right ]$
Solution: Using the formula
$\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}} \right ]$
Substituting the value we get,
$\begin{aligned} &\cos ^{-1}\left[\sin ^{-1}\left(\frac{3}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}+\frac{5}{13} \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right)\right] \\ &=\cos ^{-1}\left[\sin ^{-1}\left(\frac{3}{5} \times \frac{12}{13}+\frac{5}{13} \times \frac{4}{5}\right)\right] \\ &=\cos ^{-1}\left[\sin ^{-1}\left(\frac{36}{65}+\frac{20}{65}\right)\right] \\ &=\cos ^{-1}\left[\sin ^{-1}\left(\frac{56}{65}\right)\right] \end{aligned}$
Again, we know that
$\sin^{-1}x= \cos^{-1}\sqrt{1-x^{2}}$
Now, substituting we get
$\begin{aligned} &=\cos \left[\cos ^{-1} \sqrt{1-\left(\frac{56}{65}\right)^{2}}\right] \\ &=\cos \left[\cos ^{-1} \sqrt{\left(\frac{33}{65}\right)}\right] \end{aligned}$
$= \frac{33}{65}\; \; \; \; \; \; \; \; \; \left [ \because \cos \left ( \cos^{-1} \right )= x \right ]$
Hence $\cos \left ( \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13} \right )= \frac{33}{65}$

Inverse Trigonometric Function Exercise 3.12 Question 2 (i)

To Prove:$\sin^{-1}\frac{63}{65}= \sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
Hint: Using R.H.S term, here first we will we will convert $\sin^{-1}x$ and $\cos^{-1}x$ then use formula.
$\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}\right ]$
Solution: Taking R.H.S
R.H.S = $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
=$= \sin^{-1}\frac{5}{13}+\sin^{-1}\frac{4}{5} \; \; \; \; \; \; \; \; \; [\because \cos^{-1}x=\sin^{-1}\sqrt{1-x^{2}}]$
Then we will use the formula
$\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}\right ]$
$\begin{aligned} &= \sin ^{-1}\left[\frac{5}{13} \sqrt{1-\left(\frac{4}{5}\right)^{2}}+\frac{4}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}\right] \\ &= \sin ^{-1}\left(\frac{5}{13} \times \frac{3}{5}+\frac{4}{5} \times \frac{12}{13}\right) \\ &= \sin ^{-1}\left(\frac{15}{65}+\frac{48}{65}\right) \end{aligned}$
$= \sin^{-1}\left ( \frac{63}{65} \right )$
$= L.H.S$
Hence $\sin^{-1}\frac{63}{65}= \sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$

Inverse Trigonometric Function Exercise 3.12 Question 2 (ii) .

To prove:$\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}= \tan^{-1}\frac{63}{16}$
Hint: First we will convert $\sin^{-1}x$ and $\cos^{-1}x$ then we will use the formula of
$\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}} \right ]$
Solution: Taking L.H.S
$\begin{aligned} \text { L.H.S } &=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5} \\ &=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \sqrt{1-\left(\frac{3}{5}\right)^{2}} \quad\left[\because \cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}}\right] \\ &=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{4}{5} \end{aligned}$
Now, using the formula we have,
$\begin{aligned} \text { L. } H . S &=\sin ^{-1}\left[\frac{5}{13} \sqrt{1-\left(\frac{4}{5}\right)^{2}}+\frac{4}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}\right] \\ &=\sin ^{-1}\left(\frac{5}{13} \times \frac{3}{5}+\frac{4}{5} \times \frac{12}{13}\right) \\ &=\sin ^{-1}\left(\frac{15}{65}+\frac{48}{65}\right) \\ &=\sin ^{-1}\left(\frac{63}{65}\right) \end{aligned}$
$\begin{aligned} &=\tan ^{-1}\left(\frac{63 / 65}{1-\left(\frac{63}{65}\right)^{2}}\right) \quad\left[\sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}}\right] \\ \end{aligned}$
$= \tan^{-1}\left ( \frac{\frac{63}{65}}{\frac{16}{65}} \right )$
$\tan^{-1}\left ( \frac{63}{16} \right )= R.H.S$
Hence $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}= \tan^{-1}\frac{63}{16}$

Inverse Trigonometric Function Exercise 3.12 Question 2 (iii) .

To prove: $\frac{9\pi}{8}-\frac{9}{4}\sin ^{-1}\frac{1}{3}= \frac{9}{4}\sin^{-1}\left ( \frac{2\sqrt{2}}{3} \right )$
Hint: First we take the common in the question then we proceed.
Solution: Taking L.H.S.
L.H.S = $\frac{9 \pi}{8}-\frac{9}{4}\sin ^{-1}\frac{1}{3}$
= $\frac{9}4\left [ \frac{\pi }{2}-\sin^{-1}\frac{1}{3}\right ]$
= $\frac{9}{4}\left [ \cos^{-1}\left ( \frac{1}{3} \right ) \right ]\; \; \; \; \left [ \because \frac{\pi }{2}-\sin^{-1}x=\cos^{-1}x \right ]$
= $\frac{9}{4}\left [ \sin^{-1}\left ( 1-\sqrt{\left ( \frac{1}{3} \right )^{2}} \right ) \right ]\; \; \; \; \; \; \left [ \because \cos^{-1}x= \sin^{-1}\sqrt{1-x^{2}} \right ]$
$= \frac{9}{4}\sin^{-1}\sqrt{\frac{9-1}{9}}$
$= \frac{9}{4}\sin^{-1}\left ( \frac{2\sqrt{2}}{3} \right )$
= R.H.S
Hence,
$\frac{9\pi}{8}-\frac{9}{4}\sin ^{-1}\frac{1}{3}= \frac{9}{4}\sin^{-1}\left ( \frac{2\sqrt{2}}{3} \right )$

Inverse Trigonometric Function Exercise 3.12 Question 3 (i) .

Answer: $x= +\frac{1}{2}\sqrt{\frac{3}{7}}$
Given: $\sin^{-1}x+\sin^{-1}2x= \frac{\pi }{3}$
Hint: Using the formula $\sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$
Solution: We know that $\sin^{-1}\left ( \frac{\sqrt{3}}2{} \right )= \frac{\pi }{3}$
$\Rightarrow \sin^{-1}2x=\sin^{-1}\frac{\sqrt{3}}{2}-\sin^{-1}x$
Using the formula, $\sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$
$\Rightarrow \sin^{-1}2x= \sin^{-1}\left [ \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}-x\sqrt{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}} \right ]$
$\Rightarrow 2x=\frac{\sqrt{3}}{2}\sqrt{1-x^{2}}-\frac{x}{2}$
$\Rightarrow 2x+\frac{x}{2}= \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}$
$\Rightarrow \frac{5x}{2}= \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}$
$\Rightarrow 5x= \sqrt{3}\sqrt{1-x^{2}}$
Squaring on both sides, we get
$\Rightarrow 25x^{2}= 3\left ( 1-x^{2} \right )$
$\Rightarrow 25x^{2}= 3-3x^{2}$
$\Rightarrow 28x^{2}= 3$$x= \pm \frac{1}{2}\sqrt{\frac{3}{7}}$
As $x= - \frac{1}{2}\sqrt{\frac{3}{7}}$is not satisfying the equation
Hence $x= +\frac{1}{2}\sqrt{\frac{3}{7}}$

Inverse Trigonometric Function Exercise 3.12 Question 3 (ii) .

Answer:$x= 1$
Given:$\cos^{-1}x+\sin^{-1}\frac{x}{2}-\frac{\pi }{6}=0$
Hint: $\sin^{-1}\left (\frac{1}{2} \right )= \frac{\pi }{6}$
$\sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$
Solution: We have $\cos^{-1}x+\sin^{-1}\frac{x}{2}-\frac{\pi }{6}=0$
$\Rightarrow \left ( \frac{\pi }{2}-\sin^{-1} x\right )+\sin^{-1}\frac{x}{2}-\frac{ \pi }{6}= 0$
$\Rightarrow \sin^{-1}\frac{x}{2}-\sin^{-1}x+\frac{\pi }{2}-\frac{ \pi }{6}= 0$
$\Rightarrow \sin^{-1}\frac{x}{2}-\sin^{-1}x+\frac{\pi }{3}= 0$
$\Rightarrow \sin^{-1}\frac{x}{2}= \sin^{-1}x-\sin^{-1}\frac{\sqrt{3}}{2}\; \; \; \; \; \; \left [ \because \sin \frac{\pi }{3}= \frac{\sqrt{3}}{2} \right ]$
Using the formula,
$\sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$
$\Rightarrow \sin^{-1}\frac{x}{2}= \sin^{-1}\left [ x\sqrt{1-\left ( \frac{\sqrt{3}}{2} \right )^{2}}-\frac{\sqrt{3}}{2}\sqrt{1-x^{2}} \right ]$
$\Rightarrow \sin^{-1}\frac{x}{2}= \sin^{-1}\left [ x\sqrt{1-\left ( \frac{3}{4} \right )}-\frac{\sqrt{3}}{2}\sqrt{1-x^{2}} \right ]$
$\Rightarrow \frac{x}{2}= \frac{x}{2}-\frac{\sqrt{3}}{2}\sqrt{1-x^{2}}$
$\Rightarrow \sqrt{1-x^{2}}= 0$
Squaring on both sides, we get
$\Rightarrow 1-x^{2}= 0$
$\Rightarrow x= \pm 1$ [ As $x=-1$ is not satisfying the equation]
Hence $x=1$is the required answer.

Inverse Trigonometric Function Exercise 3.12 Question 3 (iii) .

Answer: +13
Given:$\sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}= \frac{\pi }{2}$
Hint: Here we will use $\frac{\pi }{2}-\sin^{-1}x= \cos^{-1}x$
Solution: Here we have
$\sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}= \frac{\pi }{2}$
$\sin^{-1}\frac{5}{x}= \frac{\pi }{2}-\sin^{-1}\frac{12}{x}$
$\Rightarrow \sin^{-1}\frac{5}{x}= \cos^{-1}\frac{12}{x} \; \; \; \; \left [ \because \frac{\pi }{2}-\sin^{-1}x= \cos^{-1}x \right ]$
$\Rightarrow \sin^{-1}\frac{5}{x}=\sin^{-1}\left ( \sqrt{1-\left ( \frac{12}{x} \right )^{2}} \right ) \; \; \; \; \; \; \; \; \; \; [\cos^{-1}x=\sin^{-1}\sqrt{1-x^{2}}]$
$\Rightarrow \left ( \frac{5}{x}\right )^{2} =\sqrt{1-\left ( \frac{12}{x} \right )^{2}}$
Squaring on both sides, we get
$\Rightarrow \left ( \frac{5}{x}\right )^{2} =1-\left ( \frac{12}{x} \right )^{2}$
$\Rightarrow \left ( \frac{25}{x^{2}} \right )+\left ( \frac{144}{x^{2}} \right )= 1$
$\Rightarrow x^{2}= 169$
$\Rightarrow x=\pm 13$ [ As -13 is not satisfying the equation]
Hence $x= 13$ is the result

The 3rd chapter in mathematics, Inverse Trigonometry for Class 12, has the concepts like solving sine, cosine, tangent, cotangent, secant, and cosecant. In addition, the students will be asked to find the angle of various trigonometry ratios. Exercise 3.12 contains seven questions, including the subparts. The students who find it hard to solve these questions can use the RD Sharma Class 12 Chapter 3 Exercise 3.12 solutions book.

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