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... :-)

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
$$
\frac{\binom{2n}{3}}{\binom{2n}{1}} =\frac{(2n-1)(n-1)}{3}
$$