A) \[\tan 3A\tan 2A\tan A\]
B) \[-\tan 3A\tan 2A\tan A\]
C) \[\tan A\tan 2A-\tan 2A\tan 3A-\tan 3A\tan A\]
D) None of these
Correct Answer: A
Solution :
Since \[\tan \,\,3A=\frac{\tan A+\tan 2A}{1-\tan A\,\,\tan 2A}\] \[\Rightarrow \,\,\tan \,\,3A-\tan \,\,2A-\tan A=\tan \,\,3A\,\tan \,\,2A\,\,\tan A\].You need to login to perform this action.
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