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Willow Pratt 2022-07-03 Answered
prove that $\sum _{k=1}^{n=90}\frac{1}{\mathrm{sin}\left(k-1\right)\mathrm{sin}\left(k\right)}=\frac{\mathrm{cos}{1}^{\circ }}{{\mathrm{sin}}^{2}{1}^{\circ }}$
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## Answers (1)

cefflid6y
Answered 2022-07-04 Author has 13 answers
Hint
$\begin{array}{rl}\frac{1}{\mathrm{sin}{k}^{\circ }\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(k+1{\right)}^{\circ }}& =\frac{\mathrm{sin}{1}^{\circ }}{\mathrm{sin}{1}^{\circ }\mathrm{sin}{k}^{\circ }\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(k+1{\right)}^{\circ }}\\ & =\frac{\mathrm{sin}\left(\left(k+1{\right)}^{\circ }-{k}^{\circ }\right)}{\mathrm{sin}{1}^{\circ }\mathrm{sin}{k}^{\circ }\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(k+1{\right)}^{\circ }}\\ & =\frac{1}{\mathrm{sin}{1}^{\circ }}\phantom{\rule{thinmathspace}{0ex}}\left[\mathrm{cot}{k}^{\circ }-\mathrm{cot}\left(k+1{\right)}^{\circ }\right]\end{array}$
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