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