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