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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 2 Maths Textbook Solution.

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Answer: \sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1

Hint: First we will convert \cos ^{-1} \frac{1-x^{2}}{1+x^{2}} into \tan ^{-1}

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



\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}

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

=\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\tan ^{-1} \frac{2 x}{1-x^{2}}\right\}                                 \left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]

\begin{aligned} &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-\left(\frac{1-x^{2}}{2 x}\right)\left(\frac{2 x}{1-x^{2}}\right)}\right\} \\ &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-1}\right\} \\ &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{0}\right\} \end{aligned}                                      \left(\frac{a}{0}=\infty\right)

=\sin \left\{\tan ^{-1}(\infty)\right\}                                                                   \left(\begin{array}{l} \because \tan \frac{\pi}{2}=\infty \\ \tan ^{-1}(\infty)=\frac{\pi}{2} \end{array}\right)

\begin{aligned} &=\sin \left(\frac{\pi}{2}\right) \\ &=1 \end{aligned}

Hence it is proved that \sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1


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