[MNR 1986]
A) $ 2\sec 2x $
B) $ 2\text{cosec }2x $
C) $ \sec 2x $
D) $ cosec2x $
Correct Answer: B
$ \frac{d}{dx}(\log \tan x)=\frac{1}{\tan x}{{\sec }^{2}}x=\frac{\cos x}{{{\cos }^{2}}x\sin x} $
$ =\frac{2}{2}\frac{1}{\cos x\sin x}=2cosec2x $ .
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