#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.2 question 68

Hint: you must know the rules of derivative of logarithm and trigonometric functions.

Given:   $y=\frac{1}{2} \log \left(\frac{1-\cos 2 x}{1+\cos 2 x}\right)$

Prove: $\frac{d y}{d x}=2 \operatorname{cosec} 2 x$

Solution:

$y=\frac{1}{2} \log \left(\frac{1-\cos 2 x}{1+\cos 2 x}\right)$                    $\left[\therefore 1-\cos 2 x=2 \sin ^{2} x ; 1+\cos 2 x=2 \cos ^{2} x\right]$

\begin{aligned} &y=\frac{1}{2} \log \left(\frac{2 \sin ^{2} x}{2 \cos ^{2} x}\right) \\\\ &y=\frac{1}{2} \log \tan ^{2} x \end{aligned}

\begin{aligned} &y=\frac{1}{2} \times 2 \log \tan x \\\\ &y=\log \tan x \end{aligned}

Differentiate with respect to x,use chain rule

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}(\log \tan x) \\\\ &\frac{d y}{d x}=\frac{1}{\tan x} \frac{d}{d x}(\tan x) \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{1}{\tan x} \times \sec ^{2} x \\\\ &\frac{d y}{d x}=\frac{\cos x}{\sin x} \times \frac{1}{\cos ^{2} x} \\\\ &\frac{d y}{d x}=\frac{1}{\sin x \cos x} \end{aligned}

Multiply and divide by 2

$\frac{d y}{d x}=\frac{2}{2 \sin x \cos x}$                            $[\therefore 2 \sin x \cos x=\sin 2 x]$

$\frac{d y}{d x}=\frac{2}{\sin 2 x}$                $\left[\therefore \frac{1}{\sin x}=\operatorname{cosec} x\right]$

$\frac{d y}{d x}=2 \operatorname{cosec} 2 x$

∴ Proved