Question

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

Answer 1

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

$\frac{dy}{dx}=\frac{1}{2}\frac{(1+\cos 2x)}{(1-\cos 2x)}\left[\right(1+\cos 2x\left)\right(2\sin 2x)-(1-\cos 2x\left)\right(-2\sin 2x\left)\right](1+\cos 2x{)}^2$

$\frac{dy}{dx}=\frac{1}{2}\frac{1}{(1-{\cos }^{2}2x)}(2\sin 2x+2\sin 2x)$

$\frac{dy}{dx}=\frac{2\sin 2x}{1-{\cos }^{2}x}$

$\frac{dy}{dx}=\frac{2\sin 2x}{2{\sin }^{2}2x}$

$\frac{dy}{dx}=2\csc 2x$