Trigonometric identity needed for sums involving secants I am looking for a closed-form formula for the following sum:
$\displaystyle \sum_{k=0}^{N}{\frac{\sin^{2}(\frac{k\pi}{N})}{a \cdot \sin^{2}(\frac{k\pi}{N})+1}}=\sum_{k=0}^{N}{\frac{1}{a+\csc^{2}(\frac{k\pi}{N})}}$.
Is such a formula known?

This may not be of much (any!) help, but Mathematica 7 gives a closed-form solution in terms of QPolyGamma functions:
$\frac{\psi _{e^{-\frac{2 i \pi }{n}}}^{(0)}\left(1-\frac{\log
   \left(\frac{a-2 \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i
   \pi }{n}}\right)}\right)-\psi _{e^{-\frac{2 i \pi
   }{n}}}^{(0)}\left(n-\frac{\log \left(\frac{a-2
   \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i \pi
   }{n}}\right)}+1\right)+\sqrt{a+1} n \log \left(e^{-\frac{2 i \pi
   }{n}}\right)}{a \sqrt{a+1} \log \left(e^{-\frac{2 i \pi
   }{n}}\right)}$
$+$
$\frac{\psi _{e^{-\frac{2 i \pi }{n}}}^{(0)}\left(n-\frac{\log
   \left(\frac{a+2 \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i
   \pi }{n}}\right)}+1\right)-\psi _{e^{-\frac{2 i \pi
   }{n}}}^{(0)}\left(1-\frac{\log \left(\frac{a+2
   \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i \pi
   }{n}}\right)}\right)}{a \sqrt{a+1} \log \left(e^{-\frac{2 i \pi
   }{n}}\right)}$
$\psi^{(0)}_q$ is the q-PolyGamma function. 
