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Mathematics · Trigonometry Ratios and Identities

JEE Advanced 2025 — Paper 2 — Question 31

Let α=1sin⁡60∘sin⁡61∘+1sin⁡62∘sin⁡63∘+…+1sin⁡118∘sin⁡119∘. \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 (cosec⁡1∘α)2\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^{2} is \qquad

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

α=∑r=30591sin⁡(2r)∘sin⁡(2r+1)∘\alpha=\sum_{\mathrm{r}=30}^{59} \frac{1}{\sin (2 \mathrm{r})^{\circ} \sin (2 \mathrm{r}+1)^{\circ}}

αcosec⁡1∘=∑r=3059sin⁡1∘sin⁡(2r)∘sin⁡(2r+1)∘=∑r=3059(cot⁡(2r)∘−cot⁡(2r+1)∘)=cot⁡60∘−cot⁡61∘+cot⁡62∘−cot⁡63∘⋮+cot⁡(118∘)−cot⁡(119∘)αcosec⁡1∘=cot⁡60∘=13(cosec⁡1∘α)2=3\begin{aligned} & \frac{\alpha}{\operatorname{cosec} 1^{\circ}}=\sum_{\mathrm{r}=30}^{59} \frac{\sin 1^{\circ}}{\sin (2 \mathrm{r})^{\circ} \sin (2 \mathrm{r}+1)^{\circ}} \\ & =\sum_{\mathrm{r}=30}^{59}\left(\cot (2 \mathrm{r})^{\circ}-\cot (2 \mathrm{r}+1)^{\circ}\right) \\ & =\cot 60^{\circ}-\cot 61^{\circ} \\ & \quad+\cot 62^{\circ}-\cot 63^{\circ} \\ & \quad \vdots \\ & +\cot \left(118^{\circ}\right)-\cot \left(119^{\circ}\right) \\ & \frac{\alpha}{\operatorname{cosec} 1^{\circ}}=\cot 60^{\circ}=\frac{1}{\sqrt{3}} \\ & \left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^{2}=3 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 2
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Continued Sum or Product of Series of Trigonometric Ratios