[MP PET 1986]
A) $ \cot ,2x $
B) $ 2{{\cot }^{2}}x $
C) $ 2\cot ,2x $
D) $ {{\cot }^{2}},2x $
Correct Answer: C
$ \cot x-\tan x=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{\sin x,\cos x} $ $ =\frac{2,\cos ,2x}{\sin ,2x}=2\cot 2x. $
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