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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.

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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]


\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



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