Mathematics - Differentiation Question with Solution | TestHub

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

As we know that, $\sqrt{1-\sin x} =\left|\sin \frac{x}{2}- \cos \frac{x}{2}\right| \& \sqrt{1+\sin x} =\left|\sin \frac{x}{2}+ \cos \frac{x}{2}\right|$

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

$={\cot }^{-1}\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\text{ for }x\in \left(\frac{\pi }{2},\pi \right)$

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

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

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

$y\left(x\right)=\frac{\pi }{2}-\frac{x}{2}$

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