A) \[\tan A\]
B) \[\cot A\]
C) \[\tan 2A\]
D) \[\cot 2A\]
Correct Answer: D
Solution :
\[\frac{\sin 3A-\cos \left( \frac{\pi }{2}-A \right)}{\cos A+\cos (\pi +3A)}=\frac{\sin 3A-\sin A}{\cos A-\cos 3A}\] =\[\frac{2\cos 2A\sin A}{2\sin 2A\sin A}=\frac{\cos 2A}{\sin 2A}=\cot 2A\].
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