Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=\frac{1}{2}\log \left(\frac{1-\cos 2x}{1+\cos 2x}\right) \\ \Rightarrow y=\frac{1}{2}\log \left(\frac{2{\sin }^{2}x}{2{\cos }^{2}x}\right) \\ \Rightarrow y=\frac{1}{2}\log \left({\tan }^{2}x\right) \\ \Rightarrow y=\frac{2}{2}\log \tan x \\ \Rightarrow y=\log \tan x \end{gathered}$$
Differentiate with respect to x,
$$\begin{gathered} \frac{\mathit{d}y}{\mathit{d}x}=\left(\log \tan x\right) \\ =\frac{1}{\tan x}\times \frac{d}{dx}\left(\tan x\right) \\ =\frac{{\sec }^{2}x}{\tan x} \\ =\frac{1}{{\cos }^{2}x\times {\displaystyle \frac{\sin x}{\cos x}}} \\ =\frac{1}{\sin x\cos x} \\ =\frac{2}{2\sin x\cos x} \\ =\frac{2}{\sin 2x} \\ \mathrm{So},\frac{\mathit{d}y}{\mathit{d}x}=2\mathrm{cosec}2x \end{gathered}$$