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Prove that $$\frac{1}{\sin^{2}\frac{\pi }{4k+2}}+\frac{1}{\sin^{2}\frac{3\pi }{4k+2}}+\frac{1}{\sin^{2}\frac{5\pi }{4k+2}}+\cdots+\frac{1}{\sin^{2}\frac{(2k-1)\pi }{4k+2}}=2k(k+1)$$

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Using $(7)$ from this answer, we get $$ \begin{align} \sum_{k=1}^n\frac1{\sin^2\left(\frac{2k-1}{4n+2}\pi\right)} &=\sum_{k=1}^n\csc^2\left(\frac{2k-1}{2n+1}\frac\pi2\right)\tag{1}\\ &=\sum_{k=1}^n\sec^2\left(\frac{2n-2k+2}{2n+1}\frac\pi2\right)\tag{2}\\ &=\sum_{k=1}^n\sec^2\left(\frac{k}{2n+1}\pi\right)\tag{3}\\ &=n+\sum_{k=1}^n\tan^2\left(\frac{k}{2n+1}\pi\right)\tag{4}\\[4pt] &=n+n(2n+1)\tag{5}\\[12pt] &=2n(n+1)\tag{6} \end{align} $$ Explanation:
$(1)$: $\frac1{\sin(x)}=\csc(x)$
$(2)$: $\csc(x)=\sec\left(\frac\pi2-x\right)$
$(3)$: substitute $k\mapsto n+1-k$
$(4)$: $\sec^2(x)=1+\tan^2(x)$
$(5)$: use $(7)$ from this answer
$(6)$: simplify

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  • $\begingroup$ Would the downvoter care to comment? $\endgroup$ – robjohn Aug 15 '15 at 1:43

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