(C) We know that $\cot x = \frac{\cos x}{\sin x}$ and $\tan x = \frac{\sin x}{\cos x}$.$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x

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(C) We know that $\cot x = \frac{\cos x}{\sin x}$ and $\tan x = \frac{\sin x}{\cos x}$.$\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}$Using the identities $\cos^2 x - \sin^2 x = \cos 2x$ and $2\sin x \cos x = \sin 2x$,we get:$= \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2\cos 2x}{\sin 2x} = 2\cot 2x$.