Prove $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}$,where $x \in\left(0, \frac{\pi}{4}\right)$.

Let $y = \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$.
Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+\sin x}+\sqrt{1-\sin x})$:
$y = \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^2}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$
$y = \frac{(1+\sin x) + (1-\sin x) + 2\sqrt{(1+\sin x)(1-\sin x)}}{(1+\sin x) - (1-\sin x)}$
$y = \frac{2 + 2\sqrt{1-\sin^2 x}}{2\sin x} = \frac{2 + 2\cos x}{2\sin x} = \frac{1+\cos x}{\sin x}$
Using trigonometric identities $1+\cos x = 2\cos^2 \frac{x}{2}$ and $\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2}$:
$y = \frac{2\cos^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \cot \frac{x}{2}$
Therefore,$\cot^{-1}(y) = \cot^{-1}(\cot \frac{x}{2}) = \frac{x}{2}$,which is the $R.H.S.$