Question

Find the general solution of TanA+Tan2A+√3TanA.Tan2A=√3

Answer 1

Hi ,

tanA + tan2A + √3 tanAtan2A = √3

tanA + tan2A = √3 - √3tanAtan2A

= √3 ( 1 - tanAtan2A )

( tanA + tan2A ) / ( 1 - tanAtan2A ) = √3

tan ( A + 2A ) = tan 60°

tan 3A = tan60°

3A= 60°

A = 60° /3

A = 20°

I hope this helps you.

:)