<p>The sum <math>\frac{1}{\sin 45° \sin 46°} + \frac{1}{\sin 47° \sin 48°} + \ldots + \frac{1}{\sin 133° \sin 134°}</math> is equal to</p>
Step-by-Step Solution
Key Concept: Use the identity for reciprocal of product of sines to create a telescoping series that cancels intermediate terms.
<p><strong>Step 1:</strong> For each term, use the identity: <math>\frac{1}{\sin A \sin B} = \frac{\sin(B-A)}{\sin(B-A)\sin A \sin B} = \frac{1}{\sin(B-A)}[\cot A - \cot B]</math></p><p><strong>Step 2:</strong> First term: <math>\frac{1}{\sin 45° \sin 46°} = \frac{1}{\sin 1°}[\cot 45° - \cot 46°]</math></p><p><strong>Step 3:</strong> Second term: <math>\frac{1}{\sin 47° \sin 48°} = \frac{1}{\sin 1°}[\cot 47° - \cot 48°]</math></p><p><strong>Step 4:</strong> Last term: <math>\frac{1}{\sin 133° \sin 134°} = \frac{1}{\sin 1°}[\cot 133° - \cot 134°]</math></p><p><strong>Step 5:</strong> Adding all terms, this is a telescoping series. The sum becomes: <math>\frac{1}{\sin 1°}[\cot 45° - \cot 134°]</math></p><p><strong>Step 6:</strong> Since <math>\cot 134° = \cot(180° - 46°) = -\cot 46° = -\cot(45° + 1°)</math>, and using <math>\cot 45° = 1</math>, the result simplifies to <math>\frac{1}{\sin 1°} = \cosec(1°)</math></p><p>∴ Answer is (b) cosec(1°)</p>
Correct Answer: B