Question

Evaluate: $$ \frac{1}{\sin 1^\circ \sin 2^\circ} + \frac{1}{\sin 2^\circ \sin 3^\circ} + \cdots + \frac{1}{\sin 89^\circ \sin 90^\circ} $$

• A. \( \frac{\cos 1^\circ}{\sin 1^\circ} \)
• B. \( \frac{\cos 1^\circ}{\sin^2 1^\circ} \)
• C. \( \frac{\sin 1^\circ}{\cos 1^\circ} \)
• D. \( \frac{\sin^2 1^\circ}{\cos 1^\circ} \)

Hint:

Use telescoping sums when you see symmetric reciprocal products like \( \frac{1}{\sin A \sin B} \). Look for identities to collapse the series.

Answer 1

The Correct Option is B

We are given a telescoping sum: \[ \sum_{k = 1}^{89} \frac{1}{\sin k^\circ \sin (k+1)^\circ} \] We use the identity: \[ \frac{1}{\sin A \sin B} = \frac{\cos(A - B) - \cos(A + B)}{2 \sin A \sin B} \] Instead, an efficient approach is: Let us observe that: \[ \frac{1}{\sin k^\circ \sin (k+1)^\circ} = \cot k^\circ - \cot(k+1)^\circ \] So: \[ \sum_{k=1}^{89} \frac{1}{\sin k^\circ \sin(k+1)^\circ} = \sum_{k=1}^{89} (\cot k^\circ - \cot(k+1)^\circ) \] This is a telescoping sum: \[ (\cot 1^\circ - \cot 2^\circ) + (\cot 2^\circ - \cot 3^\circ) + \cdots + (\cot 89^\circ - \cot 90^\circ) \Rightarrow \cot 1^\circ - \cot 90^\circ = \cot 1^\circ - 0 = \cot 1^\circ \] Now, \[ \cot 1^\circ = \frac{\cos 1^\circ}{\sin 1^\circ} \Rightarrow \frac{1}{\sin 1^\circ} \cdot \cot 1^\circ = \frac{1}{\sin 1^\circ} \cdot \frac{\cos 1^\circ}{\sin 1^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ} \] % Final Answer: \[ \boxed{ \frac{\cos 1^\circ}{\sin^2 1^\circ} } \]