Trigonometric Identities Question 61
Question: $ \tan 3A-\tan 2A-\tan A= $
[MNR 1982; Pb. CET 1991]
Options:
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
Show Answer
Answer:
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 $ .
BETA
