Strategy to find the sum

\[\sum_{k=45^{\circ}}^{133^{\circ}}\frac{1}{\sin(k)\sin(k+1^{\circ})}\]

part of solution

let\[\sin(x)=\frac{e^{ix}-\frac{1}{e^{ix}}}{2}\]

then \[\frac{1}{\sin(k)\sin(k+1^{\circ})}=\frac{4e^{(2k+1)i}}{(e^{2ki}-1)(e^{(2k+2)i}-1)}\]\[=\left(\frac{1}{e^{2ki}-1}-\frac{1}{e^{(2k+2)i}-1}\right)\cdot\frac{1}{e^i(e^{2i}-1)}\]