Let
$$\alpha = \dfrac{1}{\sin 60^\circ \sin 61^\circ} + \dfrac{1}{\sin 62^\circ \sin 63^\circ} + \dots + \dfrac{1}{\sin 118^\circ \sin 119^\circ}.$$\n\nThen the value of\n\n$$\left(\dfrac{\csc 1^\circ}{\alpha}\right)^2$$\n\nis
Step-by-Step Solution
Key Concept: Using the identity $\dfrac{1}{\sin A \sin B} = \dfrac{\cot A - \cot B}{\sin(B-A)}$ and utilizing the symmetry of the cotangent function about $90^\circ$ to telescopically pair and cancel terms.
Use the identity $\dfrac{\sin(B - A)}{\sin A \sin B} = \cot A - \cot B$ with difference $B - A = 1^\circ$:
$$\dfrac{\sin 1^\circ}{\sin \theta \sin(\theta + 1^\circ)} = \cot \theta - \cot(\theta + 1^\circ)$$
Applying this to all terms in $\alpha$:
$$\alpha \sin 1^\circ = (\cot 60^\circ - \cot 61^\circ) + (\cot 62^\circ - \cot 63^\circ) + \dots + (cot 118^\circ - \cot 119^\circ)$$
Rearranging the terms into even and odd cotangent groups:
$$\alpha \sin 1^\circ = \sum_{k=30}^{59} \cot(2k)^\circ - \sum_{k=30}^{59} \cot(2k+1)^\circ$$
Use the property $\cot(180^\circ - x) = -\cot x$:
- For the odd terms: $\sum_{k=30}^{59} \dots \cot(2k+1)^\circ = \cot 61^\circ + \cot 63^\circ + \dots + \cot 119^\circ = 0$ (all terms cancel by pairing $x$ and $180^\circ - x$, e.g., $\cot 61^\circ + \cot 119^\circ = 0$).
- For the even terms: $\sum_{k=30}^{59} \cot(2k)^\circ = \cot 60^\circ + \cot 62^\circ + \dots + \cot 118^\circ$.
Similarly, even terms cancel in pairs from $62^\circ$ to $118^\circ$ (with $\cot 90^\circ = 0$). Only $\cot 60^\circ$ remains.
Thus:
$$\alpha \sin 1^\circ = \cot 60^\circ = \dfrac{1}{\sqrt{3}} \implies \alpha = \dfrac{\csc 1^\circ}{\sqrt{3}}$$
Hence:
$$\left(\dfrac{\csc 1^\circ}{\alpha}\right)^2 = (\sqrt{3})^2 = 3$$
Correct Answer: