$\frac{d}{dx}\left\{{\tan }^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}=$

$\frac{d}{dx}\left\{{\tan }^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$

$=\frac{d}{dx}{\tan }^{-1}(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\times \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}})$

$=\frac{d}{dx}{\tan }^{-1}\left(\frac{1+\sin x+1-\sin x+2\sqrt{1-{\sin }^{2}x}}{1+\sin x-1+\sin x}\right)$

$=\frac{d}{dx}{\tan }^{-1}\left(\frac{2+2\cos x}{2\sin x}\right)$

$\Rightarrow \frac{d}{dx}{\tan }^{-1}\left(\frac{1+\cos x}{\sin x}\right)=\frac{d}{dx}{\tan }^{-1}\left(\frac{2{\cos }^2\frac{x}{2}}{2\sin \frac{x}{2}.\cos \frac{x}{2}}\right)$

$\Rightarrow \frac{d}{dx}.{\tan }^{-1}\cot \left(\frac{x}{2}\right).=\frac{d}{dx}{\tan }^{-1}\tan (\frac{\pi }{2}-\frac{x}{2})$

$\Rightarrow \frac{d}{dx}(\frac{\pi }{2}-\frac{x}{2})=\frac{-1}{2}$