Question:
tan 3A − tan 2A − tan A =
(a) tan 3 A tan 2 A tan A
(b) −tan 3 A tan 2 A tan A
(c) tan A tan 2 A − tan 2 A tan 3 A − tan 3 A tan A
(d) None of these
Solution:
(a) tan 3 A tan 2 A tan A
$3 A=2 A+A$
$\Rightarrow \tan 3 A=\tan (2 A+A)$
$=\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}$
$\Rightarrow \tan 3 A-\tan 3 A \tan 2 A \tan A=\tan 2 A+\tan A$
$\Rightarrow \tan 3 A-\tan 2 A-\tan A=\tan 3 A \tan 2 A \tan A$
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