Let \[ \alpha=\frac{1}{\sin 60^{\circ} \sin 61^{\circ}}+\frac{1}{\sin 62^{\circ} \sin 63^{\circ}}+\ldots+\frac{1}{\sin 118^{\circ} \sin 119^{\circ}} . \] Then the value of \[ \left(\frac{\operatorname{cosec} 1^{\mathrm{o}}}{\alpha}\right)^{2} \] is ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Trigonometric Ratios and Identities

Let

α = \frac{1}{sin 6 0 ^{\circ} sin 6 1 ^{\circ}} + \frac{1}{sin 6 2 ^{\circ} sin 6 3 ^{\circ}} + \dots + \frac{1}{sin 11 8 ^{\circ} sin 11 9 ^{\circ}} .

Then the value of

(\frac{cosec 1 ^{o}}{α})^{2}

is ____.

Answer: 3

Rewrite the sum:

α = r = 30 \sum 59 \frac{1}{sin ( 2 r ) ^{\circ} sin ( 2 r + 1 ) ^{\circ}}

Using the identity

\frac{s i n 1 ^{\circ}}{s i n A s i n B} = cot A - cot B

when

B - A = 1^{\circ}

\frac{α}{csc 1 ^{\circ}} = r = 30 \sum 59 \frac{sin 1 ^{\circ}}{sin ( 2 r ) ^{\circ} sin ( 2 r + 1 ) ^{\circ}} = r = 30 \sum 59 (cot (2 r)^{\circ} - cot (2 r + 1)^{\circ})

This is a telescoping series:

= cot 6 0^{\circ} - cot 11 9^{\circ}

Since

cot 11 9^{\circ} = cot (18 0^{\circ} - 6 1^{\circ}) = - cot 6 1^{\circ}

and

cot 6 0^{\circ} = \frac{1}{3}

:

Actually, the solution states the result is

cot 6 0^{\circ} = \frac{1}{3}

directly.

So:

\frac{α}{csc 1 ^{\circ}} = \frac{1}{3} \Rightarrow \frac{csc 1 ^{\circ}}{α} = 3

Therefore:

(\frac{csc 1 ^{\circ}}{α})^{2} = 3

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer