JEE Advanced · Mathematics · 7. Trigonometry
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 __________
- A 3
- B 5
- C 9
- D 10
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \alpha=\sum_{\mathrm{r}-30}^{59} \frac{1}{\sin (2 \mathrm{r})^{\circ} \sin (2 \mathrm{r}+1)^{\circ}} \\ & \frac{\alpha}{\operatorname{cosec} 1^{\circ}}=\sum_{\mathrm{r}-30}^{59} \frac{\sin 1^{\circ}}{\sin (2 \mathrm{r})^{\circ} \sin (2 \mathrm{r}+1)^{\circ}} \\ & =\sum_{\mathrm{r}-30}^{59}\left(\cot (2 \mathrm{r})^{\circ}-\cot (2 \mathrm{r}+1)^{\circ}\right) \\ & =\cot 60^{\circ}-\cot 61^{\circ} \\ & \quad+\cot 62^{\circ}-\cot 63^{\circ} \\ & \quad \vdots \\ & +\cot \left(118^{\circ}\right)-\cot \left(119^{\circ}\right) \\ & \frac{\alpha}{\operatorname{cosec} 1^{\circ}}=\cot 60^{\circ}=\frac{1}{\sqrt{3}} \\ & \left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2=3\end{aligned}\)
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