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

So in one of my recent post, I mentioned of one rare but extremely useful trigonometry identity that sounds:

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

We should then use it when appropriate.

In the previous blog post that asked us to prove $\tan^2 20^\circ+\tan^2 40^\circ+\tan^2 80^\circ=33$, we see the opportunity has presented itself for us to employ this trigonometric identity and expect to get a shortcut solution to this IMO contest problem.

Wow, Oh My..., that is really straightforward application of the identity:

When $x=20^{\circ}$, we have:

$\tan^2 20^{\circ}+\tan^2 (20+60^{\circ})+\tan^2 (60^{\circ}-20)=9\tan^2 3(20^{\circ})+6$

$\tan^2 20^{\circ}+\tan^2 80^{\circ}+\tan^2 40^{\circ}=9\tan^2 60^{\circ}+6$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=9(\sqrt{3})^2+6$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=27+6$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=33$

And we are hence done.

Can you see it now that some trigonometric identity can be of great help, if you remember it and use it for the right problem.