Is this a trigonometric identity? $$ \sum_{k=1}^n \frac 2 {1 - \cos\left( \frac{2k-1} n \cdot\pi \right)} = n^2 $$
If I'm not mistaken, I can prove this when $n$ is a power of $2$, by induction on the exponent. Numerical evidence suggests it is true of positive integers $n$ in general. Can it be proved?
We need $$\sum_{k=1}^n\csc^2\dfrac{(2k-1)\pi}{2n}=n+\sum_{k=1}^n\cot^2\dfrac{(2k-1)\pi}{2n}$$
Now using Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$
$$\cot2nx=\dfrac{\binom{2n}0c^{2n}-\binom{2n}2c^{2n-2}+\cdots}{\binom{2n}1c^{2n-1}+\cdots}$$ where $c=\cot x$
Now if $\cot2nx=0,2nx=(2m+1)\dfrac\pi2$ where $m$ is any integer
$$x=\dfrac{(2m+1)\pi}{2n}$$ where $m=0,1,2\cdots,2n-1$
SO, the roots of $$\binom{2n}0c^{2n}-\binom{2n}2c^{2n-2}+\cdots=0$$ are $\cot\dfrac{(2m+1)\pi}{2n};m=0,1,2\cdots,2n-1$
So, the roots of $$\binom{2n}0d^n-\binom{2n}2d^{n-1}+\cdots=0$$ are $\cot^2\dfrac{(2m+1)\pi}{2n};m=0,1,2\cdots,n-1$
$$\implies\sum_{k=1}^n\cot^2\dfrac{(2k-1)\pi}{2n}=\dfrac{\binom{2n}2}{\binom{2n}0}$$
Have I missed soemthing?