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

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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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