<s>[INST] Evaluating $\frac{\sum_{k=0}^6 \csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}$ The question is to evaluate $$\frac{\sum_{k=0}^{6}\csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}$$ where $a=\pi/8$ without looking at the trigonometric table.
I tried to transform the $\csc^2$ term to $\cot^2$ term and use addition formula but it made the problem too cumbersome.I also tried to manipulate the numerator in the form so that it telescopes but couldnot succeed.I am not in need of a full solution.Any idea or hint to proceed shall be highly appreciated.Thanks.
 [/INST] The result of the summation is:
$\frac{\sum_{k=0}^{6}\csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}=7$.
The summation, after some calculation, is reduced to:
$\frac{(A E F+B D F+C D E)+D E F}{7 D E F}=7$
where
$A=4-2\sqrt{2}+(2\sqrt{2}-2)cos(\frac{\pi}{7})$,
$B=-4+2\sqrt{2}+(2-2\sqrt{2})sin(\frac{\pi}{14})$,
$C=-4+2\sqrt{2}+(-2+2\sqrt{2})sin(\frac{3\pi}{14})$,
$D=2\sqrt{2} cos(\frac{\pi}{7})+ sin(\frac{3\pi}{14})+2$,
$E= cos(\frac{\pi}{7})- 2\sqrt{2} sin(\frac{\pi}{14})-2$,
$F=\sqrt{2} sin(\frac{3\pi}{14})+ sin(\frac{\pi}{14})-2$.
By substituting and grouping we get:
$\frac{6+\frac{1}{8}}{7 \frac{1}{8}}=7$.
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