Question

Q.32. Differentiate with respect to x
$\frac{1+\sin x}{1-\sin x}$

Answer 1

$$\begin{gathered} Lety=\frac{1+\sin x}{1-\sin x} \\ \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) \\ Applyquotientrule \\ \frac{dy}{dx}=\left(\frac{\left(1-\sin x\right)\frac{d}{dx}\left(1+\sin x\right)-\left(1+\sin x\right)\frac{d}{dx}\left(1-\sin x\right)}{{\left(1-\sin x\right)}^2}\right) \\ \frac{dy}{dx}=\left(\frac{\left(1-\sin x\right)\left(\cos x\right)-\left(1+\sin x\right)\left(-\cos x\right)}{{\left(1-\sin x\right)}^2}\right) \\ \frac{dy}{dx}=\frac{2\cos x}{{\left(1-\sin x\right)}^2} \end{gathered}$$