#### Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 1  Maths Textbook Solution.

Answer: $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$

Hint: First we will convert $\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$ into $\tan ^{-1}$

Given: $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$

Explanation:

L.H.S:

$\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$

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

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

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

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

$=\frac{\pi}{2}$

Hence it is Proved that $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$