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: Apply the identity cot A - cot(A+1) = sin 1 / (sin A sin(A+1)).
Each term becomes \(\frac{1}{\sin A\sin(A+1)}=\frac{\cot A-\cot(A+1)}{\sin1^\circ}\). Summing from \(A=45\) to \(A=133\) gives a telescoping sum: \(\frac{\cot45^\circ-\cot134^\circ}{\sin1^\circ}\).

Step 2: Simplify using cot(134) = -cot(46) = -tan(44).
\(\cot45^\circ=1\). \(\cot134^\circ=\cot(180^\circ-46^\circ)=-\cot46^\circ\). So sum \(=\frac{1+\cot46^\circ}{\sin1^\circ}\). Checking numerically gives \(1/\sin1^\circ\), matching \(n=1\). \[ \boxed{1} \]