If x is in the $1^{st}$ Quadrant, then the value of $tan^{-1}\begin{bmatrix}\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}-\sqrt{1-cosx}}\end{bmatrix}$ is :

$\frac{\pi }{4}+\frac{x}{2}$

The correct answer is Option (3) → $\frac{\pi }{4}+\frac{x}{2}$

$\tan^{-1}\left[\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right]$ $0≤x<∞$

as $\{1+\cos x=2\cos^2\frac{x}{2},1-\cos x=2\sin^2\frac{x}{2}\}$

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

$⇒\tan^{-1}\left[\frac{\sin(\frac{π}{4}+\frac{x}{2})}{\cos(\frac{π}{4}+\frac{x}{2})}\right]$

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

$=\frac{\pi }{4}+\frac{x}{2}$