Question

tanA+tan(60°+A)+tan(120°+A)=3tan3A prove the following

Answer 1

apply tan(A+B) formula i.e tanA+tanB /1 -tanAtanB

tan A + root3 + tanA /1-root3tanA + tan 120 + tanA / 1- tan120tanA

tan(180-60) = -tan60 = -root3

tanA + [root3 + tana / 1-root3tanA ]+ [(-root3) +tanA / 1+root3tanA ]

on taking the l.c.m we get

tanA(1- 3tan 2A) +(root3tanA +1)(root3 + tanA) +(tanA -root3)(1-root3tanA )/1-3tan2A

after solving brackets we get

9tanA - 3tan3A /1-3tan2A

3(3tanA - tan3A)/1 -3tan2 A

i.e 3tan3A

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