Question 15

The general term is $T_k = \frac{1}{\sin(60+2k)^\circ \sin(61+2k)^\circ}$ for $k=0$ to $29$.

Multiply by $\sin 1^\circ$:

$\sin 1^\circ T_k = \frac{\sin((61+2k) - (60+2k))}{\sin(60+2k) \sin(61+2k)} = \cot(60+2k)^\circ - \cot(61+2k)^\circ$.

Sum $\alpha \sin 1^\circ = \sum_{k=0}^{29} (\cot(60+2k)^\circ - \cot(61+2k)^\circ)$.

Using $\cot(180-\theta) = -\cot \theta$, we pair terms from the start and end:

For $k=0$: $\cot 60^\circ - \cot 61^\circ$.

For $k=29$: $\cot 118^\circ - \cot 119^\circ = -\cot 62^\circ + \cot 61^\circ$.

Summing these pairs: $(\cot 60^\circ - \cot 61^\circ) + (\cot 61^\circ - \cot 62^\circ) = \cot 60^\circ - \cot 62^\circ$.

Summing all 15 such pairs telescopes to $\cot 60^\circ - \cot 90^\circ = \frac{1}{\sqrt{3}} - 0 = \frac{1}{\sqrt{3}}$.

Thus, $\alpha \sin 1^\circ = \frac{1}{\sqrt{3}} \implies \frac{\csc 1^\circ}{\alpha} = \sqrt{3}$. $(\sqrt{3})^2 = 3$.