Question

tan x +tan (π/3+x)+tan (2π/3+x) =3​

Answer 1

Your question is -------------If tan x +tan(x+π/3)+tan(x+2π/3)=3

prove that tan 3x=1

we know:$tan (A+B)=[{tanA+tanB}/{1-tanA-tanB}]$

therefore,

$⇒tanx+{tanx+ tanπ/3}/{1-tanx-tanπ/3}+{tanx+tan2π/3}/{1-tanx-tan2π/3}=3$

$⇒tanx+{tanx+√3}/{1-√3tanx}+{tanx-√3}{1+√3tanx}=3$

$⇒tanx+(tanx+√3)(1+√3tanx)+(tanx-√3)(1-√3tanx)/(1-√3tanx)(1+√3tanx)=3$

$⇒tanx+{tanx+√3tan²x+√3+3tanx+tanx-√3tan²x-√3+3tanx}/{1-3tan²x}=3$

$⇒tanx+{8tanx}/{1-3tan²x}=3$

$⇒{tanx-3tan²x+8tanx}/{1-3tan²x}=3$

$⇒{9tanx-3tan²x}/{1-3tan²x}=3$

$⇒3{tanx-3tan²x}/{1-3tan²x}=3$

We know:$tan3A={tanA-3tan²A}/{1-3tan²A}$

therefore,$tan3x=3/3$

$tan3x=1$

$hence proved$

Answer 2

Answer:

see the attachment for the solution...