1. Tardigrade
  2. Question
  3. Mathematics
  4. The value of tan -1 (√1+x2+√1-x2/√1+x2-√1-x2) is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The value of $\tan ^{-1} \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ is equal to

Inverse Trigonometric Functions

Solution:

We have, $\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$
Let $x^2=\cos 2 \theta$. Then, we get
Given expression $=\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 \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right) $
$ \left[\because 1+\cos 2 \theta=2 \cos ^2 \theta \text { and } 1-\cos 2 \theta=2 \sin ^2 \theta\right] $
$=\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)$
(divide numerator and denominator inside the bracket by $\cos \theta$ )
$=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)$
$=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\left(\tan \frac{\pi}{4}\right)(\tan \theta)}\right)=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\theta\right)\right]$
$\left(\because \tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}\right)$
$=\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2$
$\left(\because x^2=\cos 2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^2\right)$