prove that $\sum_{k=1}^{n=90}\frac{1}{\sin(k-1)\sin(k)} =\frac{\cos1^{\circ}}{\sin^21^{\circ}}$

Question

how do I prove that $$\sum_{k=1}^{n=90}\frac{1}{\sin(k-1)\sin(k)} =\frac{\cos1^{\circ}}{\sin^21^{\circ}}$$
or $${1\over \sin1^{\circ}\sin2^{\circ}}+{1\over \sin2^{\circ}\sin3^{\circ}}+{1\over \sin3^{\circ}\sin4^{\circ}}+...+{1\over \sin89^{\circ}\sin90^{\circ}} = \frac{\cos1^{\circ}}{\sin^21^{\circ}}$$
Any ideas about how to go about doing this ?

Answer 1

Hint

\begin{align*} \frac{1}{\sin k^{\circ} \, \sin (k+1)^{\circ}} & =\frac{\sin 1^{\circ}}{\sin 1^{\circ}\sin k^{\circ} \, \sin (k+1)^{\circ}}\\ & =\frac{\sin ((k+1)^{\circ}-k^{\circ})}{\sin 1^{\circ}\sin k^{\circ} \, \sin (k+1)^{\circ}}\\ & = \frac{1}{\sin 1^{\circ}} \, \left[\cot k^{\circ}-\cot (k+1)^{\circ}\right] \end{align*}