The differential coefficient of log (tan x) is :

$\frac{d}{d x} \log (\tan x)=\frac{1}{\tan x} . \sec ^2 x$

$=\frac{\cos x}{\sin x} . \frac{1}{\cos x}=\frac{1}{\sin x \cos x}$

$=\frac{2}{2 \sin x \cos x}=\frac{2}{\sin 2 x}=2 ~cosec ~2 x$

Hence (2) is correct answer.