Solveeit Logo
Login

Question

Question: \[\tan 3A - \tan 2A - \tan A =\]...

tan⁡3A−tan⁡2A−tan⁡A=\tan 3A - \tan 2A - \tan A =

A

tan⁡3Atan⁡2Atan⁡A\tan 3A\tan 2A\tan A

B

−tan⁡3Atan⁡2Atan⁡A- \tan 3A\tan 2A\tan A

C

tan⁡Atan⁡2A−tan⁡2Atan⁡3A−tan⁡3Atan⁡A\tan A\tan 2A - \tan 2A\tan 3A - \tan 3A\tan A

D

None of these

Answer

tan⁡3Atan⁡2Atan⁡A\tan 3A\tan 2A\tan A

Explanation

Solution

Since tan⁡3A=tan⁡A+tan⁡2A1−tan⁡Atan⁡2A\tan{}3A = \frac{\tan A + \tan 2A}{1 - \tan A\tan 2A}

⇒tan⁡3A−tan⁡2A−tan⁡A=tan⁡3Atan⁡2Atan⁡A\Rightarrow \tan{}3A - \tan{}2A - \tan A = \tan{}3A\tan{}2A\tan A.