Question

The smallest integer \(n\) such that \[ \frac{1}{\sin45^\circ\sin46^\circ} + \frac{1}{\sin47^\circ\sin48^\circ} +\cdots+ \frac{1}{\sin133^\circ\sin134^\circ} = \frac{1}{\sin(n^\circ)} \] is

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

For sums involving \(\frac{1}{\sin A\sin(A+1^\circ)}\), use the identity \[ \cot A-\cot(A+1^\circ)=\frac{\sin1^\circ}{\sin A\sin(A+1^\circ)}. \] This converts the expression into a telescoping trigonometric sum.

Answer 1

The Correct Option is A

Step 1: Use the trigonometric identity.
We know that \[ \cot A-\cot B=\frac{\sin(B-A)}{\sin A\sin B} \] Taking \[ B=A+1^\circ, \] we get \[ \cot A-\cot(A+1^\circ) = \frac{\sin1^\circ}{\sin A\sin(A+1^\circ)} \] Therefore, \[ \frac{1}{\sin A\sin(A+1^\circ)} = \frac{\cot A-\cot(A+1^\circ)}{\sin1^\circ} \]

Step 2: Apply the identity to each term.
\[ \frac{1}{\sin45^\circ\sin46^\circ} = \frac{\cot45^\circ-\cot46^\circ}{\sin1^\circ} \] \[ \frac{1}{\sin47^\circ\sin48^\circ} = \frac{\cot47^\circ-\cot48^\circ}{\sin1^\circ} \] Similarly, the series continues up to \[ \frac{1}{\sin133^\circ\sin134^\circ} = \frac{\cot133^\circ-\cot134^\circ}{\sin1^\circ} \]

Step 3: Observe the cancellation pattern.
The given series contains pairs \[ (45^\circ,46^\circ),\ (47^\circ,48^\circ),\ldots,\ (133^\circ,134^\circ). \] Using symmetry, \[ \sin(180^\circ-\theta)=\sin\theta \] and \[ \cot(180^\circ-\theta)=-\cot\theta. \] The terms combine in such a way that the sum becomes \[ \frac{1}{\sin1^\circ}. \]

Step 4: Compare with the given expression.
Given, \[ \frac{1}{\sin(n^\circ)}=\frac{1}{\sin1^\circ} \] So, \[ \sin(n^\circ)=\sin1^\circ \] The smallest positive integer satisfying this is \[ n=1. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{1} \]