Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$ 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:

$\begin{matrix} 
\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}
 &amp; 
\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}
&amp;
\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}
&amp;
\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}
&amp;
\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}
 \\ 
\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}
 &amp;
 \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}
&amp;
\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}
&amp;
\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}
&amp;
\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}
\\
\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}
&amp;
\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}
&amp;
\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}
&amp;
\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}
&amp;
\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}
\\
\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}
&amp;
\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}
&amp;
\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}
&amp;
\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}
&amp;
\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}
\\
\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}
&amp;
\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}
&amp;
\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}
&amp;
\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}
&amp;
\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}
\\
\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}
&amp;
\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}
&amp;
\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}
&amp;
\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}
&amp;
\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}
 \end{matrix}$

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?
Micah already pointed the way in a comment: The identity

$$\tan a \tan b = \frac{\tan a - \tan b}{\tan(a-b)} - 1$$

causes the inner sum to telescope. To make full use of this, let's extend the inner sum to $k=2n$:

$$
\begin{align}
\sum _{l=1}^{n}\sum _{k=1}^{2n}\tan \frac {lk\pi } {2n+1}\tan \frac {l(k+1) \pi } {2n+1}
&amp;=
\sum _{l=1}^{n}\sum _{k=1}^{2n}\left(\frac{\tan \frac {l(k+1)\pi } {2n+1}-\tan \frac {lk \pi } {2n+1}}{\tan\frac {l\pi}{2n+1}}-1\right)
\\
&amp;=
\sum _{l=1}^{n}\left(\frac{\tan \frac {l(2n+1)\pi } {2n+1}-\tan \frac {l\pi } {2n+1}}{\tan\frac {l\pi}{2n+1}}-2n\right)
\\
&amp;=
\sum _{l=1}^{n}(0-1-2n)
\\
&amp;=
-n(2n+1)\;.
\end{align}
$$

This sum contains each of the terms we want to sum twice, with mirror symmetry, and it contains one additional term in the middle for $k=n$. Thus, for our sum to vanish, we need

$$
\sum _{l=1}^{n}\tan \frac {ln\pi } {2n+1}\tan \frac {l(n+1) \pi } {2n+1}=-n(2n+1)\;.
$$

The arguments of the two factors add to $l\pi$, so they're negatives of each other, so we're looking for

$$
-\sum _{l=1}^{n}\tan^2 \frac {ln\pi } {2n+1}\;.
$$

We can again extend the sum to $2n$ to double it, since the arguments form pairs that add up to $n\pi$; then, since $n$ and $2n+1$ are coprime, we can replace $ln$ by $l$ while traversing the same arguments; and then we can set the upper limit back to $n$, since the arguments still add up to $\pi$ in pairs. Thus, what we need is 

$$
\sum _{l=1}^{n}\tan^2 \frac {l\pi } {2n+1}=n(2n+1)\;.
$$

How to find this sum is shown at Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$. Adapting the argument in the accepted answer there for our odd denominator, we obtain

$$
\left(\cos\frac{k\pi}{2n+1}+\mathrm i\sin\frac{k\pi}{2n+1}\right)^{2n+1}=(-1)^k\;,
$$

taking the imaginary part,

$$
\sum_{r=0}^n\binom{2n+1}{2r+1}\left(\cos\frac{k\pi}{2n+1}\right)^{2n-2r}\left(\mathrm i\sin\frac{k\pi}{2n+1}\right)^{2r+1}=0\;,
$$

dividing by $\left(\cos\frac{k\pi}{2n+1}\right)^{2n+1}$,

$$
\sum_{r=0}^n\binom{2n+1}{2r+1}\left(\mathrm i\tan\frac{k\pi}{2n+1}\right)^{2r+1}=0\;,
$$

and dividing by $\tan\frac{k\pi}{2n+1}$ and letting $x=\tan^2\frac{k\pi}{2n+1}$,

$$
\sum_{r=0}^n\binom{2n+1}{2r+1}(-x)^r=0\;.
$$

Then Vieta's formula shows that the sum of the roots of this equation is

$$\frac{\binom{2n+1}2}{\binom{2n+1}0}=n(2n+1)\;,$$

as required.

[Update:]

This answer suggests an alternative, more elementary way to calculate the sum of the squares of the tangents: On the grid $\frac{l\pi}{2n+1}$, the tangent decomposes into $n$ mutually orthogonal sines, each of whose dot product with itself is $2n+1$, so the dot product of the tangent with itself is $n(2n+1)$.
To prove that,

$$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$

Consider the following,

$$\tan ( \frac {l(k+1)\pi } {2n+1} - \frac {l(k)\pi } {2n+1} ) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}}$$

i.e. $$\tan ( \frac {l\pi } {2n+1} ) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}}$$

i.e. $$\ ( 1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {\tan \frac {l\pi } {2n+1}}$$

i.e. $$\tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1} = (\frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {\tan \frac {l\pi } {2n+1}} - 1)$$

Hold l = 1 and take $$\sum _{k=1}^{n-1}$$

Next hold l = 2 and take $$\sum _{k=1}^{n-1}$$

....... 

Continue this till l exhausts for l = n

Now sum all the terms you get and we see that the sum is indeed equal to 0

i.e. $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$