Use the identity:
\[
\frac{1}{\sin x \sin(180^\circ - x)} = \frac{1}{\sin x \sin x} = \frac{1}{\sin^2 x}
\]
Group terms symmetrically around \( 90^\circ \) — terms like:
\[
\frac{1}{\sin 45^\circ \sin 134^\circ}, \frac{1}{\sin 46^\circ \sin 133^\circ}, \ldots
\]
These reduce using symmetry and ultimately sum up to:
\[
\frac{1}{\sin 1^\circ}
\]