If $$x \in \left[ {\frac{\pi }{2},\pi} \right]$$   then $${\cot ^{ - 1}}\left( {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right) = $$

Solution :
$$\eqalign{ & \sqrt {1 + \sin x} = \sin \frac{x}{2} + \cos \frac{x}{2} \cr & \sqrt {1 - \sin x} = \sin \frac{x}{2} - \cos \frac{x}{2} \cr & \left[ {{\text{for}}\,\frac{\pi }{4} \leqslant \frac{x}{2} \leqslant \frac{\pi }{2}\sin \frac{x}{2} \geqslant \cos \frac{x}{2}} \right] \cr & \therefore {\text{the expression is}} \cr & {\cot ^{ - 1}}\left( {\frac{{\sin \frac{x}{2} + \cos \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2}}}{{\sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2} + \cos \frac{x}{2}}}} \right) \cr & = {\cot ^{ - 1}}\left( {\tan \frac{x}{2}} \right) = {\cot ^{ - 1}}\cot \left( {\frac{\pi }{2} - \frac{x }{2}} \right) = \frac{{\pi - x}}{2} \cr} $$