Evaluate Trigonometric Series Sum and Cosecant Ratio Squared

To evaluate the given sum, we multiply both sides of the expression for $\alpha$ by $\sin 1^\circ$:

$\sin 1^\circ \alpha = \frac{\sin(61^\circ - 60^\circ)}{\sin 60^\circ \sin 61^\circ} + \frac{\sin(63^\circ - 62^\circ)}{\sin 62^\circ \sin 63^\circ} + \dots + \frac{\sin(119^\circ - 118^\circ)}{\sin 118^\circ \sin 119^\circ}$

Using the identity $\frac{\sin(y-x)}{\sin x \sin y} = \cot x - \cot y$, we rewrite the series as:

$\sin 1^\circ \alpha = (\cot 60^\circ - \cot 61^\circ) + (\cot 62^\circ - \cot 63^\circ) + \dots + (\cot 118^\circ - \cot 119^\circ)$

Grouping the positive and negative terms separately:

$\sin 1^\circ \alpha = \left( \sum_{k=30}^{59} \cot(2k^\circ) \right) - \left( \sum_{k=30}^{59} \cot((2k+1)^\circ) \right)$

By symmetric cancellation using $\cot(180^\circ - \theta) = -\cot \theta$, all pairs in the second sum cancel to zero, and in the first sum all terms cancel except for $\cot 60^\circ$:

$\sin 1^\circ \alpha = \cot 60^\circ = \frac{1}{\sqrt{3}}$

Rearranging gives:

$\frac{\csc 1^\circ}{\alpha} = \sqrt{3}$

Squaring both sides yields:

$\left( \frac{\csc 1^\circ}{\alpha} \right)^2 = 3$