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Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise Multiple Choice Questions Question 31 maths Textbook Solution.

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Answer: \frac{2 a}{1-a^{2}}

Hint: Try to separate \sin ^{-1} function and \cos ^{-1} function into RHS, so that the variables get free.

Given: \sin ^{-1}\left(\frac{2 a}{1-a^{2}}\right)+\cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)


\sin ^{-1}\left(\frac{2 a}{1-a^{2}}\right)+\cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)

\begin{aligned} &\Rightarrow 2 \tan ^{-1} a+2 \tan ^{-1} a=2 \tan ^{-1} x \\ &\Rightarrow 4 \tan ^{-1} a=2 \tan ^{-1} x \\ &\Rightarrow 2 \tan ^{-1} a=\tan ^{-1} x \end{aligned}

\tan ^{-1}\left(\frac{2 a}{1-a^{2}}\right)=\tan ^{-1} x

x=\frac{2 a}{1-a^{2}}

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