\[\frac{d}{dx}[{{\tan }^{-1}}(\cot x)+{{\cot }^{-1}}(\tan x)]=\]

1.	\[\frac{d}{dx}[{{\tan }^{-1}}(\cot x)+{{\cot }^{-1}}(\tan x)]=\]
A.	0
B.	1
C.	? 1
D.	? 2
Answer» E.

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\[\frac{d}{dx}[{{\tan }^{-1}}(\cot x)+{{\cot }^{-1}}(\tan x)]=\]

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