If tan x + tan(x+π/3) + tan(x+2π/3) = 3, Prove that (3tanx − tan³x)/(1 − 3tan²x) = −1

Question

If

\[ \tan x+\tan\left(x+\frac{\pi}{3}\right)+\tan\left(x+\frac{2\pi}{3}\right)=3 \]

prove that:

\[ \frac{3\tan x-\tan^3x} {1-3\tan^2x} =-1 \]

Proof

Using

\[ \tan3x = \frac{3\tan x-\tan^3x} {1-3\tan^2x} \]

it is enough to prove that:

\[ \tan3x=-1 \]

Let

\[ \tan x=t \]

Then,

\[ t+\frac{t+\sqrt3}{1-\sqrt3t} +\frac{t-\sqrt3}{1+\sqrt3t} =3 \]

Taking LCM,

\[ t(1-3t^2) +(t+\sqrt3)(1+\sqrt3t) +(t-\sqrt3)(1-\sqrt3t) \]

\[ = 3(1-3t^2) \]

\[ t-3t^3+t+3t^2+\sqrt3+ t-3t^2+\sqrt3 \]

\[ +t-3t^2-\sqrt3+t+3t^2-\sqrt3 = 3-9t^2 \]

\[ 3t-3t^3 = 3-9t^2 \]

\[ 3t-t^3 = 1-3t^2 \]

\[ \frac{3t-t^3}{1-3t^2} =1 \]

Hence,

\[ \frac{3\tan x-\tan^3x} {1-3\tan^2x} =-1 \]

Hence proved.