Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$ How can you prove that
$$
\sum_{k=1}^{n-1}\tan^{2}\left(\frac{k \pi}{2n}\right)
=
\frac{\left(n-1\right)\left(2n - 1\right)}{3}
$$
for every integer $n\geq 1$ > ?.
PS: no, it's not a homework... :-)
 A: By a well know formula we have
$$
\left(\cos{\frac{k\pi}{2n}} + i\sin{\frac{k\pi}{2n}}\right)^{2n}=(-1)^{k}
$$
Hence by Binomial theorem we have ($[x]$ is not an integer part, brackets are added for clarity)
$$
\sum_{t=0}^{2n}\binom{2n}{t}\left[\cos{\frac{k\pi}{2n}}\right]^t \cdot \left[i\sin{\frac{k\pi}{2n}}\right]^{2n-t}=(-1)^{k}
$$
Now we consider only imaginary part of this:
$$
\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left[\cos{\frac{k\pi}{2n}}\right]^{2r+1} \cdot \left[i\sin{\frac{k\pi}{2n}}\right]^{2n-2r-1}=0
$$
Divide it by $[\cos{\frac{k\pi}{2n}}]^{2n}$:
$$
\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left[i\tan{\frac{k\pi}{2n}}\right]^{2n-2r-1}=0
$$
Now multiply by $i\tan{\frac{k\pi}{2n}}$:
$$
\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left[i\tan{\frac{k\pi}{2n}}\right]^{2n-2r}=0
$$
So $[\tan{\frac{k\pi}{2n}}]^2$ are roots of the following polynomial:
$$
\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left[-x\right]^{n-r}=0
$$
Hence by Vieta's_formulas sum of it roots is equal to
$$
\frac{\binom{2n}{3}}{\binom{2n}{1}} =\frac{(2n-1)(n-1)}{3}
$$
A: I am doing a similar thing for $\cot$ i hope you can reciprocate it.
We have $$(\cot(\theta)-i)^{n} = \frac{ \cos{n\theta} - \sin{n \theta}}{\sin^{n}(\theta)}$$
Equating the real and imaginary parts on both sides we have $$ \frac{\sin(n\theta)}{\sin{\theta}} = \sum\limits_{s} {n \choose 2s+1}  (-1)^{s} \cot^{n-2s-1}(\theta)$$
Now take $2n+1$ instead of $n$ we have, $$\sin{(2n+1)\theta} = \sin^{2n+1}(\theta)P_{n}\cot^{2}(\theta)$$
for $\displaystyle 0 < \theta < \frac{\pi}{2}$ and where $P_{n}$ is the polynomial given by $${ 2n+1 \choose 1}T^{n} - {2n+1 \choose 3}T^{n-1} + \cdots$$
Noting that the zeros of $P_{n}$ are precisely $\displaystyle \frac{r\pi}{2n+1},\ n =1,2,..$, we have the first identity from the sum of the roots formula. 
From the inequality $\sin{x} < x < \tan{x}$ we have $$\cot^{2m}(x) < \frac{1}{x^{2m}} < (1 + \cot^{2}(x))^{m}$$ we have $$ \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1}  < \frac{(2n+1)^2m}{\pi^{2m}} \sum\limits_{r=1}^{n} \frac{1}{r^{2m}} < \sum\limits_{r=1}^{n} \Bigl( 1 + \cot^{2} \frac{r\pi}{2n+1} \Bigr)^{m}$$
Therefore, $$\sum\limits_{r=1}^{n} \Bigl(1+\cot^{2} \frac{r\pi}{2n+1} \Bigr)^{m} =  \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} + \mathcal{O}(n^{2m-1})$$
In other words to find $c_{2m}$ where $ \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} = c_{2m}n^{2m} + \mathcal{O}(n^{2m-1})$ it suffices to look at the sum, $$ \cot^{2m} \frac{\pi}{2n+1} + \cot^{2m} \frac{2\pi}{2n+1} + \cdots + cot^{2m} \frac{n\pi}{2n+1}$$
which is the sum $s_{m}$ of $m-th$ powers of the roots of the polynomial $P_{n}$. Then you can complete it using newtons formula for the sum of the roots.
A: This problem is over 3 years old, but I was just made aware of it, so I thought I would add a new method of attack, similar to this answer.
Here we use the function $\dfrac{2n/z}{z^{2n}-1}$ which has residue $1$ at each $2n^\text{th}$ root of unity and residue $-2n$ at $0$.
Since $\tan^2\left(\dfrac\theta2\right)=-\left(\dfrac{z-1}{z+1}\right)^2$ for $z=e^{i\theta}$, define
$$
f(z)=-\left(\frac{z-1}{z+1}\right)^2\frac{2n/z}{z^{2n}-1}
$$
The sum of the residues of $f$ is $0$ since $|f(z)|\sim2n/z^{2n+1}$ as $|z|\to\infty$ and so an integral around a large circle vanishes.
On the other hand, the sum of the residues is twice the sum we want plus the residue at $0$ and the residue at $-1$.
$\hspace{32mm}$
Thus,
$$
\begin{align}
\sum_{k=1}^{n-1}\tan^2\left(\frac{k\pi}{2n}\right)
&=-\frac12\left(\operatorname*{Res}_{z=0}(f(z))+\operatorname*{Res}_{z=-1}(f(z))\right)\\
&=-\frac12\left(2n-\frac23(2n^2+1)\right)\\
&=\frac{2n^2-3n+1}{3}
\end{align}
$$
A: I give a proof of a general case of that type of sum related to binomial coefficients in my post Tan binomial formulas from a set S and its k-subsets
