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 sum we need to evaluate.
We want $S = \displaystyle\sum_{k=1}^{89} \frac{1}{\sin k^\circ \sin(k+1)^\circ}$.
Step 2: Use the identity involving cotangent differences.
Notice that $\sin(B-A) = \sin B \cos A - \cos B \sin A$. With $B = (k+1)^\circ$ and $A = k^\circ$: \[\sin 1^\circ = \sin((k+1)^\circ - k^\circ) = \sin(k+1)^\circ\cos k^\circ - \cos(k+1)^\circ\sin k^\circ.\]
Step 3: Divide both sides by $\sin k^\circ \sin(k+1)^\circ$.
\[\frac{\sin 1^\circ}{\sin k^\circ \sin(k+1)^\circ} = \frac{\cos k^\circ}{\sin k^\circ} - \frac{\cos(k+1)^\circ}{\sin(k+1)^\circ} = \cot k^\circ - \cot(k+1)^\circ.\] So $\dfrac{1}{\sin k^\circ \sin(k+1)^\circ} = \dfrac{\cot k^\circ - \cot(k+1)^\circ}{\sin 1^\circ}$.
Step 4: Write the telescoping sum.
\[S = \frac{1}{\sin 1^\circ}\sum_{k=1}^{89}(\cot k^\circ - \cot(k+1)^\circ) = \frac{\cot 1^\circ - \cot 90^\circ}{\sin 1^\circ}.\]
Step 5: Evaluate $\cot 90^\circ$.
$\cot 90^\circ = \dfrac{\cos 90^\circ}{\sin 90^\circ} = 0$. So $S = \dfrac{\cot 1^\circ}{\sin 1^\circ} = \dfrac{\cos 1^\circ}{\sin 1^\circ \cdot \sin 1^\circ} = \dfrac{\cos 1^\circ}{\sin^2 1^\circ}$.
Step 6: State the final answer.
The sum telescopes beautifully to $\dfrac{\cos 1^\circ}{\sin^2 1^\circ}$. \[ \boxed{\dfrac{\cos 1^\circ}{\sin^2 1^\circ}} \]