# Prove that

Question:

Prove that

$\tan ^{-1} \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^{2}$

Solution:

$\tan ^{-1} \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^{2}$

Taking L.H.S,

Let $x^{2}=\cos 2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^{2}$

So, L.H.S. $=\tan ^{-1}\left[\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right]$

$=\tan ^{-1}\left[\frac{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}\right]$

$=\tan ^{-1}\left[\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right]$

$=\tan ^{-1}\left[\frac{1+\tan \theta}{1-\tan \theta}\right]$

$=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}\right]$

$=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right)=\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^{2}$

= R.H.S

– Hence Proved