If $y=\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$, show that $\frac{d y}{d x}$ is independent of $x$.
$y=\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$
Multiplying numerator and denominator
$y=\cot ^{-1}\left\{\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})}\right\}$
$y=\cot ^{-1}\left\{\frac{1+\sin x+1-\sin x+2 \sqrt{1+\sin x} \sqrt{1-\sin x}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}}\right\}$
$y=\cot ^{-1}\left\{\frac{2+2 \sqrt{1-\sin ^{2} x}}{(1+\sin x)-(1-\sin x)}\right\}$
$y=\cot ^{-1}\left\{\frac{2+2 \sqrt{1-\sin ^{2} x}}{2 \sin x}\right\}$
$y=\cot ^{-1}\left\{\frac{2(1+\cos x)}{2 \sin x}\right\}$
Using $\sin ^{2} \theta+\cos ^{2} \theta=1$
$y=\cot ^{-1}\left\{\frac{1+\cos x}{\sin x}\right\}$
Using $2 \sin \theta \cos \theta=\sin 2 \theta$ and $2 \cos ^{2} \theta-1=\cos 2 \theta$
$y=\cot ^{-1}\left\{\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right\}$
Now
$y=\cot ^{-1}\left\{\cot \frac{x}{2}\right\}$
$y=\frac{x}{2}$
Differentiating w.r.t $\mathrm{x}$, we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{2}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}$