The value of $\tan . 3 A - \tan 2 A - \tan A$ is equal to

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$\tan 3 A \tan 2 A \tan A$

none of these

$- \tan 3 A \tan 2 A \tan A$

$\tan A \tan 2 A - \tan 2 A \tan 3 A - \tan 3 A \tan A$

answer is A.

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Detailed Solution

We have

$\Rightarrow \tan 3 A = \tan (2 A + A) \Rightarrow \tan 3 A = \frac{(\tan 2 A + \tan A)}{(1 - \tan 2 A \tan A)} \Rightarrow \tan 3 A - \tan 2 A - \tan A = \tan 3 A \tan 2 A \tan A$

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