Step 1: Standard telescoping identity.

$\frac{\sin 1^{\circ }}{\sin k^{\circ }\sin (k+1{)}^{\circ }}=\cot k^{\circ }-\cot (k+1{)}^{\circ }$, using $\sin ((k+1{)}^{\circ }-k^{\circ })=\sin (k+1{)}^{\circ }\cos k^{\circ }-\cos (k+1{)}^{\circ }\sin k^{\circ }$.

Step 2: Multiply $\alpha$ by $\sin 1^{\circ }$.

$\alpha \sin 1^{\circ }=\sum _{r=30}^{59}(\cot (2r{)}^{\circ }-\cot (2r+1{)}^{\circ })$.

Step 3: Note the telescoping is not adjacent.

The pairs are $(\cot 60^{\circ }-\cot 61^{\circ })+(\cot 62^{\circ }-\cot 63^{\circ })+\cdots +(\cot 118^{\circ }-\cot 119^{\circ })$. Group using $\cot (180^{\circ }-\theta )=-\cot \theta$.

For each $\theta \in \{61^{\circ },63^{\circ },\dots ,89^{\circ }\}$, the term $-\cot \theta$ pairs with $\cot (180^{\circ }-\theta )=-\cot \theta$ from the second half, doubling cancellations. After careful pairing, all terms cancel except $\cot 60^{\circ }$.

Step 4: Conclude.

$\alpha \sin 1^{\circ }=\cot 60^{\circ }=\frac{1}{\sqrt{3}}$, so $\frac{\alpha }{\mathrm{cosec}{1}^{\circ }}=\frac{1}{\sqrt{3}}$.

Therefore ${\left(\frac{\mathrm{cosec}{1}^{\circ }}{\alpha }\right)}^2=3$.