Question

If \[ \frac{1}{\sin1^\circ\sin2^\circ} + \frac{1}{\sin2^\circ\sin3^\circ} +\cdots+ \frac{1}{\sin89^\circ\sin90^\circ} = \] then its value is

• A. \(\dfrac{\cos1^\circ}{\sin1^\circ}\)
• B. \(\dfrac{\cos1^\circ}{\sin^21^\circ}\)
• C. \(\dfrac{\sin1^\circ}{\cos1^\circ}\)
• D. \(\dfrac{\sin^21^\circ}{\cos1^\circ}\)

Hint:

For sums involving \[ \frac{1}{\sin A\sin B}, \] try converting them into cotangent differences using: \[ \cot A-\cot B= \frac{\sin(B-A)}{\sin A\sin B}. \] This usually creates a telescoping series.

Answer 1

The Correct Option is B

Step 1: Write the given series.
We have \[ S= \sum_{k=1}^{89} \frac{1}{\sin k^\circ \sin(k+1)^\circ} \] We use the identity \[ \cot A-\cot B = \frac{\sin(B-A)}{\sin A\sin B} \]

Step 2: Apply the identity.
Take \[ A=k^\circ,\qquad B=(k+1)^\circ \] Then, \[ \cot k^\circ-\cot(k+1)^\circ = \frac{\sin1^\circ}{\sin k^\circ\sin(k+1)^\circ} \] Therefore, \[ \frac{1}{\sin k^\circ\sin(k+1)^\circ} = \frac{\cot k^\circ-\cot(k+1)^\circ}{\sin1^\circ} \]

Step 3: Substitute into the series.
Hence, \[ S= \frac{1}{\sin1^\circ} \sum_{k=1}^{89} \left( \cot k^\circ-\cot(k+1)^\circ \right) \]

Step 4: Use telescoping cancellation.
Expanding, \[ S= \frac{1}{\sin1^\circ} [ \cot1^\circ-\cot2^\circ +\cot2^\circ-\cot3^\circ +\cdots +\cot89^\circ-\cot90^\circ ] \] All middle terms cancel, giving \[ S= \frac{1}{\sin1^\circ} [ \cot1^\circ-\cot90^\circ ] \] Since \[ \cot90^\circ=0, \] we get \[ S= \frac{\cot1^\circ}{\sin1^\circ} \]

Step 5: Simplify the expression.
Using \[ \cot1^\circ=\frac{\cos1^\circ}{\sin1^\circ}, \] we obtain \[ S= \frac{\cos1^\circ}{\sin^21^\circ} \]

Step 6: Final conclusion.
Therefore, \[ \boxed{ \frac{\cos1^\circ}{\sin^21^\circ} } \]