Let

$$\alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}.$$

Then the value of

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

is ______.

The given sum contains angles that differ by $$1^\circ$$, so we first convert each term by the identity
$$\frac{1}{\sin A\,\sin B}\;=\;\frac{\cot A-\cot B}{\sin\!\left(B-A\right)}\,.$$ Here $$B-A = 1^\circ$$ for every pair.

Hence each term becomes
$$\frac{1}{\sin k^\circ \,\sin(k+1)^\circ}\;=\;\frac{\cot k^\circ-\cot(k+1)^\circ}{\sin1^\circ}\,.$$ With $$k = 60,62,64,\dots ,118$$ we get

$$\alpha \;=\;\frac{1}{\sin1^\circ}\! \Bigl[\bigl(\cot60^\circ-\cot61^\circ\bigr) +\bigl(\cot62^\circ-\cot63^\circ\bigr) +\cdots +\bigl(\cot118^\circ-\cot119^\circ\bigr)\Bigr].$$

Introduce the compact notation

$$S \;=\;\sum_{k=0}^{29}\Bigl(\cot(60+2k)^\circ-\cot(61+2k)^\circ\Bigr).$$ Then $$\alpha=\dfrac{S}{\sin1^\circ}\,,$$ so our task is to evaluate $$S$$.

Separate even and odd arguments:

$$S =\bigl[\cot60^\circ+\cot62^\circ+\cdots+\cot118^\circ\bigr] -\bigl[\cot61^\circ+\cot63^\circ+\cdots+\cot119^\circ\bigr].$$

Write each angle as either below $$90^\circ$$ or above $$90^\circ$$:

• For $$\theta\lt90^\circ$$, $$\cot\theta^\circ=\tan(90^\circ-\theta^\circ).$$
• For $$\theta=90^\circ+\delta^\circ$$, $$\cot\theta^\circ=-\tan\delta^\circ.$$

Applying these to all terms gives

$$\begin{aligned} S&=\Bigl[\tan30^\circ+\tan28^\circ+\cdots+\tan2^\circ\Bigr] -\Bigl[\tan29^\circ+\tan27^\circ+\cdots+\tan1^\circ\Bigr] \\ &\quad-\Bigl[\tan28^\circ+\tan26^\circ+\cdots+\tan2^\circ\Bigr] +\Bigl[\tan29^\circ+\tan27^\circ+\cdots+\tan1^\circ\Bigr]. \end{aligned}$$

The two long odd-angle sums cancel, and every even-angle term from $$2^\circ$$ to $$28^\circ$$ cancels as well. The only term left is

$$S=\tan30^\circ=\frac{1}{\sqrt3}\,.$$

Therefore

$$\alpha=\frac{S}{\sin1^\circ}=\frac{1}{\sqrt3\,\sin1^\circ}\,.$$

Finally,

$$\left(\frac{\csc1^\circ}{\alpha}\right)^2 =\left(\frac{1/\sin1^\circ}{1/(\sqrt3\,\sin1^\circ)}\right)^2 =\left(\sqrt3\right)^2 =3.$$

Answer: 3