[MNR 1982; Pb. CET 1991]
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
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 $ .
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