This problem appears to be a fit for the telescoping technique. A natural guess is that $$\frac{1}{\\sin 1^\circ \sin 2^\circ}=\frac{1}{K}\Big(A-B\Big)$$ where $K$ is a constant and $A$ and $B$ are two trigonometric values which can form a telescoping sequence. In order to find this constant coefficient $K$, let's try to rewrite the right side to an appropriate form. $$\frac{\cos 1^\circ}{\sin^2 1^\circ} =\frac{1}{\sin 1^\circ}\cdot\cot 1^\circ$$ Given the fact $\cot 90^\circ = 0$, a conjuncture is that $$\frac{1}{\sin 1^\circ\sin 2^\circ}+\cdots+\frac{1}{\sin 89^\circ\sin 90^\circ}=\frac{1}{\sin 1^\circ}(\cot 1^\circ - \cot 90^\circ)$$ This is equivalent to showing that, for $n=1, 2, \cdots, 89$, $$\frac{1}{\sin n^\circ \sin (n+1)^\circ}=\frac{1}{\sin n^\circ}\cdot(\cot n^\circ - \cot (n+1)^\circ)$$ This indeed holds. Let's expanding the right side: \begin{align*} &\frac{1}{\sin 1^\circ}\cdot(\cot n^\circ - \cot (n+1)^\circ)\\ &=\frac{1}{\sin 1^\circ}\cdot\Big(\frac{\cos n^\circ}{\sin n^\circ}-\frac{\cos (n+1)^\circ}{\sin (n+1)^\circ}\Big)\\ &=\frac{1}{\sin 1^\circ}\cdot\frac{\cos n^\circ\sin(n+1)^\circ - \sin n^\circ\cos(n+1)^\circ}{\sin n^\circ \sin (n+1)^\circ}\\ &=\frac{1}{\sin 1^\circ}\cdot\frac{\sin 1^\circ}{\sin n^\circ \sin (n+1)^\circ}\\ &=\frac{1}{\sin n^\circ \sin (n+1)^\circ} \end{align*}