show that \[\sum\limits_{k=1}^{n-1}\csc^2\left(\dfrac{k\pi}{n}\right)=\dfrac{n^2-1}{3}\]

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show that \[\sum\limits_{k=1}^{n-1}\csc^2\left(\dfrac{k\pi}{n}\right)=\dfrac{n^2-1}{3}\]

Mathematics
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Any identity relating to products of sine? We can rearrange the sum as follows: \[ \sum_{k=1}^{n-1}\csc^2\left(\dfrac{k\pi}{n}\right)=\frac{\sum_{i=1}^{n-1}\prod_{k\neq i}^{n-1}\sin^2\left(\frac{k\pi}{n}\right)}{\prod_{k=1}^{n-1}\sin^2\left(\frac{k\pi}{n}\right)} \]

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That is, the numerator is the coefficient of the \(x\) of the polynomial \(p(x)=\prod\limits_{i=1}^{n-1}\left(x+\sin^2\left(\dfrac{k\pi}{n}\right)\right)\) and the denominator is the constant term of \(p(x)\).
\[\sum_{k=1}^{n-1} \csc^2(\frac{k\pi}{n})=\sum_{k=1}^{n-1}1+\sum_{k=1}^{n-1} \cot^2(\frac{k\pi}{n})=(n-1)+\sum_{k=1}^{n-1} \cot^2(\frac{k\pi}{n})\] \[(n-1)+\sum_{k=1}^{n-1}(\frac{1+\cos(\frac{2k\pi}{n})}{1-\cos(\frac{2k\pi}{n})})\] No idea, just messing around a bit :p
Two relevant links on Math Stack Exchange. They suggested using Chebyshev polynomial of the first and second kind to solve the problem. I cannot make this work. Maybe someone can on here. https://math.stackexchange.com/questions/122836/sum-of-the-reciprocal-of-sine-squared https://math.stackexchange.com/questions/45144/proving-frac1-sin2-frac-pi14-frac1-sin2-frac3-pi14/

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