I need help proving the following identity.


I am not sure if it is even true.


If $\tan3y=\tan30^\circ$

$3y=n180^\circ+30^\circ$ where $n$ is any integer

$y=60^\circ n+10^\circ$ where $n=0,1,2$

For $n=2,y=130^\circ,\tan130^\circ=\tan(180^\circ-50^\circ)=-\tan50^\circ$

Now $\tan3y=\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}$

and consequently, $\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}=\dfrac1{\sqrt3}$ as $\tan30^\circ=\dfrac1{\sqrt3}$

Rearrange to form a cubic equation in $\tan y$ where $y=60^\circ n+10^\circ$ where $n=0,1,2$

We need $\tan^210^\circ+\tan^250^\circ+\tan^270^\circ$


$=[\tan10^\circ+(-\tan50)+\tan70^\circ]^2$ $-2[\tan10^\circ(-\tan50)+(-\tan50)\tan70^\circ+\tan70^\circ\tan10^\circ]$


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