# Differentiate the following w.r.t x: $\tan^{-1}\Large {\bigg(\frac{\sqrt {1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\bigg)},\normalsize -1 < x < 1,x\neq 0$

Toolbox:
• $1+\cos x=2\cos^2\large\frac{x}{2}$
• $1-\cos x=2\sin^2\large\frac{x}{2}$
• $\large\frac{1-\tan x}{1+\tan x}$$=\tan\big(\large\frac{\pi}{4}$$-x)$
Step 1:
$y=\tan^{-1}\bigg[\large\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\bigg]$
Put $x^2=\cos\theta\Rightarrow \theta=\cos^{-1}x$
$\qquad\qquad=\tan^{-1}\bigg[\large\frac{\sqrt{1+\cos^2x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\bigg]$
But $1+\cos x=2\cos^2\large\frac{x}{2}$ and $1-\cos x=2\sin^2\large\frac{x}{2}$
$y=\tan^{-1}\bigg[\large\frac{\sqrt 2\cos \large\frac{x}{2}+\sqrt 2\sin\large\frac{x}{2}}{\sqrt 2\cos\Large\frac{x}{2}-\sqrt 2\sin\Large\frac{x}{2}}\bigg]$
$\;\;=\tan^{-1}\bigg[\large\frac{\cos x/2+\sin x/2}{\cos x/2-\sin x/2}\bigg]$
Step 2:
Divide each term by $\cos x/2$
$\;\;=\tan^{-1}\bigg[\large\frac{1+\tan\Large\frac{x}{2}}{1-\tan\Large\frac{x}{2}}\bigg]$