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If $\mathrm{y}=\frac{1}{2} \log \left(\frac{1-\cos 2 \mathrm{x}}{1+\cos 2 \mathrm{x}}\right)$, prove that $\frac{\mathrm{dy}}{\mathrm{dx}}=2 \operatorname{cosec} 2 \mathrm{x}$.


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

We have $1+\cos (2 \theta)=2 \cos ^{2} \theta$ and $1+\cos (2 \theta)=2 \sin ^{2} \theta$.

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

$\Rightarrow y=\frac{1}{2} \log \left(\tan ^{2} x\right)$

$\Rightarrow y=\frac{1}{2} \log (\tan x)^{2}$

$\Rightarrow y=\frac{1}{2} \times 2 \log (\tan x)\left[\because \log \left(a^{m}\right)=m \times \log (a)\right]$

$\Rightarrow y=\log (\tan x)$

On differentiating y with respect to $x$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}[\log (\tan \mathrm{x})]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\tan x} \frac{d}{d x}(\tan x)$ [using chain rule]

However, $\frac{d}{d x}(\tan x)=\sec ^{2} x$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\tan x} \times \sec ^{2} x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sec ^{2} \mathrm{x}}{\tan \mathrm{x}}$

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{\cos ^{2} x} \times \frac{\cos x}{\sin x}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sin x \cos x}$

We have $\sin (2 \theta)=2 \sin \theta \cos \theta$

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

$\Rightarrow \frac{d y}{d x}=\frac{2}{\sin 2 x}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \operatorname{cosec} 2 \mathrm{x}$

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


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