The simplest form of $\tan^{-1} \left[ \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right]$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$ ##
We have, $\tan^{-1} \left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right)$
Put $x = \cos 2\theta$, so that $\theta = \frac{1}{2} \cos^{-1} x$
$\tan^{-1} \left( \frac{\sqrt{1 + \cos 2\theta} - \sqrt{1 - \cos 2\theta}}{\sqrt{1 + \cos 2\theta} + \sqrt{1 - \cos 2\theta}} \right)$
$= \tan^{-1} \left( \frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2 \theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2 \theta}} \right)$
$= \tan^{-1} \left( \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \right) \quad \text{(Divide by } \cos \theta)$
$= \tan^{-1} \left( \frac{1 - \tan \theta}{1 + \tan \theta} \right) = \tan^{-1} \left[ \tan \left( \frac{\pi}{4} - \theta \right) \right]$
$= \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$