Question

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

Answer 1

\[\text{ We have, y} = \frac{1}{2}\log\left( \frac{1 - \cos2x}{1 + \cos2x} \right)\]

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

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

\[ \Rightarrow y = \frac{2}{2}\log \tan x \]

\[ \Rightarrow y = \log \tan x\]

Differentiate with respect to x,

\[\frac{d y}{d x} = \left( \log \tan x \right)\]

\[ = \frac{1}{\tan x} \times \frac{d}{dx}\left( \tan x \right) \]

\[ = \frac{\sec^2 x}{\tan x}\]

\[ = \frac{1}{\cos^2 x \times \frac{\sin x}{\cos x}}\]

\[ = \frac{1}{\sin x \cos x}\]

\[ = \frac{2}{2\sin x \cos x}\]

\[ = \frac{2}{\sin2x} \]

\[So, \frac{d y}{d x} = 2\text{cosec }2x\]