#### Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 3 Maths Textbook Solution.

Answer: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$

Hints: We will first convert $2 x \sqrt{1-x^{2}}$ into sin

Given: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$

$x=\sin \theta$

Let $\sqrt{1-x^{2}}=\sqrt{1-\sin ^{2} \theta}=\sqrt{\cos ^{2} \theta}=\cos \theta$                                            $\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \cos ^{2} \theta=1-\sin ^{2} \theta \end{array}\right]$

Now

\begin{aligned} &\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \\ &=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\ &=\sin ^{-1}(2 \sin \theta \cos \theta) \end{aligned}                                                                           $[2 \sin \theta \cos \theta=\sin 2 \theta]$

\begin{aligned} &=2 \theta \\ &=2 \sin ^{-1} x \end{aligned}                                                                                                            $\left[\begin{array}{l} \because \sin ^{-1}(\sin \theta)=\theta \\ x=\sin \theta \\ \theta=\sin ^{-1} x \end{array}\right]$

Hence it is proved that $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$