A) \[\tan 5x-\tan 3x-\tan 2x\]
B) \[\frac{\sin 5x-\sin 3x-\sin 2x}{\cos 5x-\cos 3x-\cos 2x}\]
C) 0
D) None of these
Correct Answer: A
Solution :
We have \[5x=3x+2x\Rightarrow \tan 5x=\tan (3x+2x)\] Þ \[\tan 5x=\frac{\tan 3x+\tan 2x}{1-\tan 3x\tan 2x}\] Þ\[\tan 5x-\tan 5x\tan 3x\tan 2x=\tan 3x+\tan 2x\] Þ \[\tan 5x\tan 3x\tan 2x=\tan 5x-\tan 3x-\tan 2x\].
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