Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$ Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$.
I tried some elementary ways while all failed.

From Sum of tangent functions where arguments are in specific arithmetic series and Prove the trigonometric identity $(35)$,
$$\tan13x=\frac{\binom{13}1t-\binom{13}3t^3+\binom{13}5t^5-\binom{13}7t^7+\binom{13}9t^9-\binom{13}{11}t^{11}+t^{13}}{\cdots}$$
where $\displaystyle t=\tan x$
Now if $\displaystyle\tan13x=0,13x=n\pi$ where $n$ is any integer 
$\displaystyle\implies x=\frac{n\pi}{13}$ where $0\le n\le13-1$
As $\displaystyle\tan0=0,\tan\frac{n\pi}{13},1\le n\le12$ are the roots of 
$\displaystyle\binom{13}1-\binom{13}3t^2+\binom{13}5t^4-\binom{13}7t^6+\binom{13}9t^8-\binom{13}{11}t^{10}+t^{12}=0\  \ \ \ (1)$
As $\displaystyle\tan(r\pi-y)=-\tan y,$ for any integer $r$
the given relation  $\displaystyle(i)\sum\limits_{k=1}^{12}\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}=2\sum\limits_{k=1}^6\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}$
$\displaystyle(ii)u_k=\tan^2\frac{k\pi}{13},1\le k\le6$
shall be the roots of 
$\displaystyle\binom{13}1-\binom{13}3u+\binom{13}5u^2-\binom{13}7u^3+\binom{13}9u^4-\binom{13}{11}u^5+u^6=0\  \ \ \  (2)$
Now, $\displaystyle y_k=\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}=\frac{3u_k-u_k^2}{1-3u_k}$ (using $\tan3A$ formula )
$\displaystyle\implies u_k^2=3u_k-y_k(1-3u_k)=3u_k(1+y_k)-y_k$
So, we need to transform Equation $\#(2)$ in terms of $y$ 
Then apply Vieta's Formulas.
But my Question is why $\displaystyle\tan\frac{k\pi}{13}$ is paired with $\displaystyle\tan\frac{3k\pi}{13},$ but not with $\displaystyle\tan\frac{2k\pi}{13}$ or $\displaystyle\tan\frac{5k\pi}{13}$ etc.?
