if the

Let $y=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$   ...(1)

Then, $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

$=\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$

$=\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1-\sin x)(1+\sin x)}}{(1+\sin x)-(1-\sin x)}$

$=\frac{2+2 \sqrt{1-\sin ^{2} x}}{2 \sin x}$

$=\frac{1+\cos x}{\sin x}$

$=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}$

$=\cot \frac{x}{2}$

Therefore, equation (1) becomes

$y=\cot ^{-1}\left(\cot \frac{x}{2}\right)$

$\Rightarrow y=\frac{x}{2}$

$\therefore \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}$