Question

TanA + Tan(60° + A ) + Tan(120°+A) =

Answer 1

You just need to expand the function
Tan A +  (tan 60 + tan A )/(1-tan60.tanA) +  (tan (120) + tan A )/(1-tan120.tanA)
tan A +  [root(3) + tanA]/[1-root(3).tanA] + [ tan A - root(3) ] /[1 + root(3).tanA]
tanA +  [ root(3) + 3.tanA + tanA + root(3).tan^2A  + tanA - root(3) -root(3).tan^2A + 3.tanA ] / [ 1 - 3tan^2A ]
tanA + [ 8tanA] / [ 1-3tan^2A ]
[9tanA - 3.tan^3A ]/[1-3tan^2A]
= 3.tan3A   ( eqaul to RHS )
Hence proved