TITLE: Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$
QUESTION [50 upvotes]: Prove that $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$


It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:
$\begin{matrix} 
\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}
 & 
\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}
&
\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}
&
\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}
&
\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}
 \\ 
\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}
 &
 \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}
&
\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}
&
\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}
&
\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}
\\
\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}
&
\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}
&
\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}
&
\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}
&
\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}
\\
\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}
&
\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}
&
\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}
&
\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}
&
\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}
\\
\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}
&
\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}
&
\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}
&
\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}
&
\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}
\\
\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}
&
\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}
&
\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}
&
\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}
&
\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}
 \end{matrix}$
one can notice then, that the first column vanish the fourth one :
$\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$
$\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$
$\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$
$\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$
$\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$
$\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$
and the third column vanish the fifth one :
$\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$
$\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$
$\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$
$\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$
$\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$
$\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$
while the second column is self-vanishing:
$\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$
$\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$
$\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .
So the equality occurs.
But how to generalize the proof?

REPLY [24 votes]: Starting with
$$
\tan(x-y) = \frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\tag{1}
$$
we get
$$
\tan(x)\tan(y)=\frac{\tan(x)-\tan(y)}{\tan(x-y)}-1\tag{2}
$$
Thus,
$$
\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right)
=\frac{\tan\left(\frac{l(k+1)\pi}{2n+1}\right)-\tan\left(\frac{lk\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-1\tag{3}
$$
Therefore, because of the telescoping sum,
$$
\begin{align}
\sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right)
&=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)-\tan\left(\frac{l\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-(n-1)\\
&=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-n\tag{4}
\end{align}
$$
Note that 
$$
\frac{\tan\left(\frac{(2n+1-l)n\pi}{2n+1}\right)}{\tan\left(\frac{(2n+1-l)\pi}{2n+1}\right)}
=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}\tag{5}
$$
so that by replacing the odd $l$s with even $2n+1-l$s and using $\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$, we get
$$
\begin{align}
\sum_{l=1}^n\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}
&=\sum_{l=1}^n\frac{\tan\left(\frac{2ln\pi}{2n+1}\right)}{\tan\left(\frac{2l\pi}{2n+1}\right)}\\
&=-\sum_{l=1}^n\frac{\tan\left(\frac{l\pi}{2n+1}\right)}{\tan\left(\frac{2l\pi}{2n+1}\right)}\\
&=\frac12\sum_{l=1}^n\left(\tan^2\left(\frac{l\pi}{2n+1}\right)-1\right)\tag{6}
\end{align}
$$

Using contour integration, we will compute
$$
\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)=n(2n+1)\tag{7}
$$
Note that
$$
\frac{(2n+1)/z}{z^{2n+1}-1}\tag{8}
$$
has simple poles. It has residue $1$ at $z=e^{\large\frac{2\pi li}{2n+1}}$ for each $l$ and residue $-(2n+1)$ at $z=0$.
Furthermore, at $z=e^{i\theta}$,
$$
-\left(\dfrac{z-1}{z+1}\right)^2=\tan^2(\theta/2)\tag{9}
$$
Because the total residue of 
$$
f(z)=\left(\frac{z-1}{z+1}\right)^2\frac{(2n+1)/z}{z^{2n+1}-1}\tag{10}
$$
is $0$, we get that the sum of its residues at $z=0$ and $z=-1$ equals
$$
\sum_{l=1}^{2n}\tan^2\left(\frac{\pi l}{2n+1}\right)=2\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)\tag{11}
$$
First,
$$
\mathrm{Res}_{z=0}f(z)=-(2n+1)\tag{12}
$$
Next,
$$
\begin{align}
\mathrm{Res}_{z=-1}f(z)
&=\mathrm{Res}_{z=0}f(z-1)\\
&=\mathrm{Res}_{z=0}(2n+1)\left(\frac{z-2}{z}\right)^2\frac1{1-z}\frac1{1+(1-z)^{2n+1}}\\
&=\mathrm{Res}_{z=0}(2n+1)\left(1-\frac4z+\frac4{z^2}\right)(1+z+\dots)\frac12\left(1+\frac{2n+1}{2}z+\dots\right)\\
&=\mathrm{Res}_{z=0}\frac{2n+1}{2}\left(1-\frac4z+\frac4{z^2}\right)\left(1+\frac{2n+3}{2}z+\dots\right)\\
&=\mathrm{Res}_{z=0}\frac{4n+2}{z^2}+\frac{(2n+1)^2}{z}+\dots\\
&=(2n+1)^2\tag{13}
\end{align}
$$
Combining $(11)$, $(12)$, and $(13)$, yields $(7)$.

Combining $(4)$, $(6)$, and $(7)$ yields
$$
\sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right)=0\tag{14}
$$