Challenging Math Contest Problem: Prove that $\tan^2 x+\tan^2 (x+60^{\circ})+\tan^2 (60^{\circ}-x)=9\tan^2 3x+6$

Given $\tan x+\tan (x+60^{\circ})-\tan (60^{\circ}-x)=3\tan 3x$,

prove that $\tan^2 x+\tan^2 (x+60^{\circ})+\tan^2 (60^{\circ}-x)=9\tan^2 3x+6$.

Wow! This is another exquisite problem that one cannot afford to pass it up but to take it as one tough learning example problem so to train students to be the best.  Remember that good teachers will forever encourage learning for understanding and are concerned with developing their students’ critical-thinking skills, problem-solving skills, and problem-approach behaviors. This problem fulfills the these goals of training students to think creatively and that is the reason I bring it to our table.

I won't beat about the bush, so here goes my plan of attack:

First notice that

i.

$\begin{align*}\small\tan 3x&=\dfrac{\tan x(3-\tan^2 x)}{1-3\tan^2x}\\&= \tan x \left(\dfrac{(\sqrt{3})^2-\tan^2 x}{1^2-(\sqrt{3} \tan x)^2} \right)\\&=\tan x \left(\dfrac{(\sqrt{3}-\tan x)(\sqrt{3}+\tan x)}{(1+\sqrt{3} \tan x)(1-\sqrt{3} \tan x)} \right)\\&=\tan x \tan(60-x) \tan (60+x)\end{align*}$

ii.

$\tan 3x=\dfrac{3\tan x-\tan^3 x}{1-3\tan^2x}$

$\rightarrow\,\,\tan^3 x+\tan3x=3\tan x+3\tan^2 x \tan 3x$

iii.

We are given $\tan x^{\circ}+\tan(60+x)^{\circ}-\tan(60-x)^{\circ}=3\tan 3x^{\circ}$

If we rewrite it as $\tan(60+x)^{\circ}-\tan(60-x)^{\circ}=3\tan 3x^{\circ}-\tan x^{\circ}$ and squaring it, we get:

$\tan^2(60+x)^{\circ}+\tan^2(60-x)^{\circ}-2\tan(60+x)^{\circ}\tan(60-x)^{\circ}$

$=9\tan^2 3x^{\circ}+\tan^2 x^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}$

Modifying it a bit we get

$\tan^2 x^{\circ}+\tan^2(60+x)^{\circ}+\tan^2(60-x)^{\circ}$

$=9\tan^2 3x^{\circ}+2\tan^2 x^{\circ}+2\tan(60+x)^{\circ}\tan(60-x)^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}$

$=9\tan^2 3x^{\circ}+2\tan^2 x^{\circ}+\dfrac{2\tan 3x^{\circ}}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}$

$=9\tan^2 3x^{\circ}+\dfrac{2(\tan^3 x^{\circ}+\tan 3x^{\circ})}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}$

$=9\tan^2 3x^{\circ}+\dfrac{2(3\tan x^{\circ}+3\tan^2 x^{\circ}\tan 3x^{\circ})}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}$

$=9\tan^2 3x^{\circ}+6+6\tan 3x^{\circ}\tan x^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}$

$=9\tan^2 3x^{\circ}+6$

This identity is nevertheless a gem that deserves to be far more widely known-and be used. Don't believe it? Then I politely invite you to solve the (Previous problem) using this very identity.