Question

If $y=ta{n}^{-1}\left(\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right)$ for $0<x<\frac{\pi }{2}$ then find $\frac{dy}{dx}$

Answer 1

$y={\tan }^{-1}\left(\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right)\to 0<x<\frac{\pi }{2}$ $\frac{dy}{dx}=?$

Lets similarity inside expression ${\tan }^{-1}\left(f\right(x\left)\right)$

$f\left(x\right)=\left(\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right)\times \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$

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

$f\left(x\right)=\frac{2\sin x}{2+2\cos x}=\frac{\sin x}{1+\cos x}$

OR $f\left(x\right)=\frac{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}-\sqrt{1-\sin x})}{(\sqrt{1+\sin x}+\sqrt{1-\sin x})(\sqrt{1+\sin x}-\sqrt{1-\sin x})}$

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

$\tan y=cosecx-\cot x$

$se{c}^{2}y\frac{dy}{dx}=-cosecx\cot x+cose{c}^{2}x$

$\frac{dy}{dx}={\cos }^{2}y\times cosecx(cosecx-cotx)$

$\frac{dy}{dx}={\cos }^{2}y\tan y.\times cosecx=\sin y\cos ycosecx$

$\frac{dy}{dx}=\frac{1}{2}\sin 2ycosecx=\frac{1}{2}\sin {\tan }^{-1}(cosecx-\cot x)cosecx$