#### 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.

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

Explanation:

L.H.S:

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

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

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

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