#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 16 maths

Answer:$\sec x(\sec x+\tan x)$

Hint: You must know the rules of solving derivative of trigonometric function.

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

Solution:

Let  $y=\sqrt{\frac{1+\sin x}{1-\sin x}}$

Differentiating with respect to x

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

$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{\frac{1}{2}-1} \times \frac{d}{d x}\left(\frac{1+\sin x}{1-\sin x}\right)$

$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{1}{2}} \times\left\{\frac{(1-\sin x)(\cos x)-(1+\sin x)(-\cos x)}{(1-\sin x)^{2}}\right\}$$\ldots \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

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

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

$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1+\sin x}(1-\sin x)^\frac{3}{2}}$

$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1+\sin x} \sqrt{1-\sin x}(1-\sin x)}$

$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1-\sin ^{2} x} \times(1-\sin x)}$

$\frac{d y}{d x}=\frac{\cos x}{\cos x(1-\sin x)}$

\begin{aligned} &\frac{d y}{d x}=\frac{1}{(1-\sin x)} \times\left(\frac{1+\sin x}{1-\sin x}\right) \\ &\frac{d y}{d x}=\frac{(1+\sin x)}{\left(1-\sin ^{2} x\right)} \end{aligned}

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

$\frac{d y}{d x}=\frac{1}{(\cos x)}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)$

$\frac{d y}{d x}=\sec x(\sec x+\tan x)$