1. **State the problem:** We need to find the value of $\left(\frac{\csc 1^\circ}{\alpha}\right)^2$ where $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ 2. **Analyze and simplify the sum $\alpha$:** Each term in the sum is $$ \frac{1}{\sin k^\circ \sin (k+1)^\circ} \, \text{for} \, k=60,62,64,\ldots,118. $$ Notice that the terms increment by 2 degrees, starting at 60. 3. **Use trigonometric identities to simplify the term:** Recall the identity $$ \csc x = \frac{1}{\sin x} $$ and the sine difference formula: $$ \sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right). $$ Try to express $$ \frac{1}{\sin k^\circ \sin (k+1)^\circ} $$ in a telescoping form. 4. **Find a telescoping expression:** Notice that $$ \frac{1}{\sin k^\circ \sin (k+1)^\circ} = \frac{\csc k^\circ \csc (k+1)^\circ}{1} $$ but this does not immediately telescope. Instead, consider rewriting using the identity $$ \frac{1}{\sin x \sin (x+1)} = \frac{\cot x - \cot (x+1)}{\sin 1^\circ}. $$ 5. **Verify the key identity:** Using the difference of cotangents, $$ \cot x - \cot (x+1) = \frac{\cos x}{\sin x} - \frac{\cos (x+1)}{\sin (x+1)} = \frac{\sin (x+1) \cos x - \sin x \cos (x+1)}{\sin x \sin (x+1)} = \frac{\sin ((x+1)-x)}{\sin x \sin (x+1)} = \frac{\sin 1^\circ}{\sin x \sin (x+1)}. $$ Rearranging: $$ \frac{1}{\sin x \sin (x+1)} = \frac{\cot x - \cot (x+1)}{\sin 1^\circ} $$ 6. **Apply the telescoping sum:** For $k=60,62,\ldots,118$, which are even degrees from 60 to 118, $$ \alpha = \sum_{k=60,62,\ldots,118} \frac{1}{\sin k^\circ \sin (k+1)^\circ} = \frac{1}{\sin 1^\circ} \sum_{k=60,62,\ldots,118} (\cot k^\circ - \cot (k+1)^\circ). $$ 7. **Rewrite the summation with telescoping:** Since terms cancel, $$ \sum (\cot k^\circ - \cot (k+1)^\circ) = \sum \cot k^\circ - \sum \cot (k+1)^\circ. $$ We pair terms and many will cancel except the first and last Check the terms carefully: Terms are for k=60,62,64,...,118. Next terms in the subtracted sum are for k+1 = 61,63,65,...,119. So the sum expands to $$ (\cot 60^\circ - \cot 61^\circ) + (\cot 62^\circ - \cot 63^\circ) + \cdots + (\cot 118^\circ - \cot 119^\circ). $$ 8. **Simplify the total sum:** Rearranging, $$ \alpha = \frac{1}{\sin 1^\circ} \bigl( (\cot 60^\circ + \cot 62^\circ + \cdots + \cot 118^\circ) - (\cot 61^\circ + \cot 63^\circ + \cdots + \cot 119^\circ) \bigr). $$ Because the angles interleave, this is the difference of cotangents of even angles from 60 to 118 and odd angles from 61 to 119. 9. **Pair terms and evaluate:** Since the range is symmetric and the difference of cotangents of angles that differ by 1 degree decreases smoothly, sum these numerically or notice a pattern: Alternatively, the sum is telescoping as initially expressed. Due to the symmetric nature, the sum simplifies further to $$ \alpha = \frac{\cot 60^\circ - \cot 119^\circ}{\sin 1^\circ}. $$ 10. **Calculate $\cot 60^\circ$ and $\cot 119^\circ$: ** - $\cot 60^\circ = \frac{1}{\sqrt{3}} \approx 0.57735$ - $\cot 119^\circ = \cot (180^\circ - 61^\circ) = -\cot 61^\circ \approx -0.55431$ 11. **Combine values:** $$ \cot 60^\circ - \cot 119^\circ = 0.57735 - (-0.55431) = 1.13166 $$ 12. **Calculate $\sin 1^\circ$:** $$ \sin 1^\circ \approx 0.017452 $$ 13. **Compute $\alpha$: ** $$ \alpha = \frac{1.13166}{0.017452} \approx 64.83 $$ 14. **Recall the original expression:** $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 = \left( \frac{1/\sin 1^\circ}{\alpha} \right)^2 = \left( \frac{1}{\sin 1^\circ \alpha} \right)^2. $$ 15. **Calculate $\csc 1^\circ=1/\sin 1^\circ \approx 57.298$: ** $$ \frac{\csc 1^\circ}{\alpha} = \frac{57.298}{64.83} \approx 0.8834 $$ Square this: $$ 0.8834^2 \approx 0.78 $$ 16. **Exact simplified result:** But from the telescoping nature and algebraic simplifications, the exact value is $$ \left(\frac{\csc 1^\circ}{\alpha}\right)^2 = \left(\frac{1}{\sin 1^\circ \alpha}\right)^2 = \left(\frac{1}{\cot 60^\circ - \cot 119^\circ}\right)^2 = \left(\frac{1}{1.13166}\right)^2 = \left(\frac{1}{\cot 60^\circ - \cot 119^\circ} \right)^2. $$ Numerically approximately $0.78$. **Final answer:** $$ \boxed{\left( \frac{\csc 1^\circ}{\alpha} \right)^2 = \left( \frac{1}{\cot 60^\circ - \cot 119^\circ} \right)^2 \approx 0.78}. $$