JEE Advanced 2025 Paper 2 · Q15 · Trigonometric Identities & Equations

Question

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 $\left(\dfrac{\csc 1^{\circ}}{\alpha}\right)^2$ is __________.

SolveKar AI · Accepted Answer

Use the identity $\frac{1}{\sin A \sin(A+1^{\circ})} = \frac{1}{\sin 1^{\circ}}(\cot A - \cot(A + 1^{\circ}))$. The sum has $30$ terms with $A = 60, 62, \ldots, 118$. $\alpha = \frac{1}{\sin 1^{\circ}} \cdot S$ where $S = \sum_{k=0}^{29} (\cot(60 + 2k) - \cot(61 + 2k)) = (\cot 60 + \cot 62 + \ldots + \cot 118) - (\cot 61 + \cot 63 + \ldots + \cot 119)$. Use $\cot(180^{\circ} - x) = -\cot x$. $B = \cot 61 + \cot 63 + \ldots + \cot 119$: pair $(61, 119), (63, 117), \ldots, (89, 91)$: each pair sums to $0$. There are $30$ odd-args from $61$ to $119$ forming $15$ such pairs, so $B = 0$. $A = \cot 60 + \cot 62 + \ldots + \cot 118$: pair $(62, 118), (64, 116), \ldots, (88, 92)$: $14$ pairs sum to $0$. Remaining: $\cot 60$ and $\cot 90 (= 0)$. So $A = \cot 60 = \frac{1}{\sqrt{3}}$. $S = A - B = \frac{1}{\sqrt{3}}$. $\alpha = \frac{1}{\sin 1^{\circ} \cdot \sqrt{3}}$. $\frac{\csc 1^{\circ}}{\alpha} = \frac{1/\sin 1^{\circ}}{1/(\sin 1^{\circ} \cdot \sqrt{3})} = \sqrt{3}$. $\left(\frac{\csc 1^{\circ}}{\alpha}\right)^2 = 3$.

JEE Advanced 2025 Paper 2 · Q15 · Trigonometric Identities & Equations

Solution