Let α= 1 60^ 61^ + 1 62^ 63^ + + 1 118^ 119^ Then the value of ( cosec 1^ α )^2 is __________

Correct answer: 0 aligned & α=∑_ r-30 ^ 59 1 (2 r)^ (2 r+1)^ & α cosec 1^ =∑_ r-30 ^ 59 1^ (2 r)^ (2 r+1)^ & =∑_ r-30 ^ 59 ( (2 r)^ - (2 r+1)^ ) & = 60^ - 61^ & + 62^ - 63^ & & + (118^ )- (119^ ) & α cosec 1^ = 60^ = 1 3 & ( cosec 1^ α )^2=3 aligned

JEE Advanced 2025Paper 2

Exam: JEE Advanced
Year: 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Advanced 2025 · Paper 2 · Mathematics · Trigonometric Ratios & Identities

Let α= 1 60^ 61^ + 1 62^ 63^ + + 1 118^ 119^ Then the value of (...

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^{\circ}}{\alpha}\right)^2

is __________

Correct answer

Step-by-step solution

\begin{aligned} & \alpha=\sum_{\mathrm{r}-30}^{59} \frac{1}{\sin (2 \mathrm{r})^{\circ} \sin (2 \mathrm{r}+1)^{\circ}} \\ & \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}

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