Question

Differentiate

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

Answer 1

$$\begin{gathered} \mathrm{Let}y=\sqrt{\frac{1+\sin x}{1-\sin x}} \\ \mathrm{Differentiate}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x\mathrm{we}\mathrm{get}, \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}{\left(\frac{1+\sin x}{1-\sin x}\right)}^{\frac{1}{2}} \\ =\frac{1}{2}{\left(\frac{1+\sin x}{1-\sin x}\right)}^{\frac{1}{2}-1}\frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) \\ =\frac{1}{2}{\left(\frac{1-\sin x}{1+\sin x}\right)}^{\frac{1}{2}}\left[\frac{\left(1-\sin x\right)\left(\cos x\right)-\left(1+\sin x\right)\left(-\cos x\right)}{{\left(1-\sin x\right)}^2}\right] \\ =\frac{1}{2}\frac{{\left(1-\sin x\right)}^{{\displaystyle \frac{1}{2}}}}{{\left(1+\sin x\right)}^{{\displaystyle \frac{1}{2}}}}\left[\frac{\cos x-\cos x\sin x+\cos x+\sin x\cos x}{{\left(1-\sin x\right)}^2}\right] \\ =\frac{1}{2}\times \frac{2\cos x}{\sqrt{1+\sin x}\left(1-\sin x\right){\displaystyle \frac{3}{2}}} \\ =\frac{\cos x}{\sqrt{1+\sin x}\left(1-\sin x\right)\frac{3}{2}} \\ =\frac{\cos x}{\sqrt{1+\sin x}\sqrt{1-\sin x}\left(1-\sin x\right)} \\ =\frac{\cos x}{\sqrt{1-{\sin }^{2}x}\times \left(1-\sin x\right)} \\ =\frac{\cos x}{\cos x\left(1-\sin x\right)}\left[\mathrm{Using}1-{\sin }^{2}x={\cos }^{2}x\right] \\ =\frac{1}{\left(1-\sin x\right)}\times \frac{\left(1+\sin x\right)}{\left(1+\sin x\right)} \\ =\frac{\left(1+\sin x\right)}{\left(1-{\sin }^{2}x\right)} \\ =\frac{1+\sin x}{{\cos }^{2}x} \\ =\frac{1}{\cos x}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right) \\ =\sec x\left(\sec x+\tan x\right) \\ \mathrm{Hence},\frac{dy}{dx}=\sec x\left(\sec x+\tan x\right) \end{gathered}$$