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Question

Mathematics Question on Trigonometric Functions

The value of tan⁡ 3A−tan⁡ 2A−tan⁡ A\tan \, 3A- \tan \, 2A- \tan \, A is:

A

tan 3A tan 2A tan A

B

tan 3A + tan 2A + tan A

C

tan 3A - tan 2A tan A

D

tan 3A + tan 2A tan A

Answer

tan 3A tan 2A tan A

Explanation

Solution

We have, 3A = 2A+ A ⇒\Rightarrow tan3A = tan(2A + A) =(tan⁡2A+tan⁡A)(1−tan⁡2Atan⁡A)= \frac{\left(\tan2 A + \tan A\right)}{\left(1- \tan2 A \tan A\right)} ⇒ \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