The differential coefficient of log (tan x) is :

2 sec 2x 2 cosec 2x 2 sec2x 2 cosec2x

2 cosec 2x

$\frac{d}{d x} \log (\tan x)=\frac{1}{\tan x} . \sec ^2 x$ $=\frac{\cos x}{\sin x} . \frac{1}{\cos x}=\frac{1}{\sin x \cos x}$ $=\frac{2}{2 \sin x \cos x}=\frac{2}{\sin 2 x}=2 ~cosec ~2 x$ Hence (2) is correct answer.