Question

The value of $\tan^{-1} \left[\frac{\sqrt{1+x^{2}} + \sqrt{1-x^{2}}}{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}\right] , \left|x\right| < \frac{1}{2}, x \ne0, $ is equal to :

• A. $\frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^{2} $
• B. $\frac{\pi}{4} + \cos^{-1} x^{2} $
• C. $\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x^{2} $
• D. $\frac{\pi}{4} - \cos^{-1} x^{2} $

Answer 1

Answer: (A)

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