Let alpha = frac{1}{sin 60^{circ} sin 61^{cir | vedclass

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(C) We are given $\alpha = \sum_{k=0}^{29} \frac{1}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}}$.Multiplying and dividing by $\sin 1^{\circ}$,we get:$\al

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) We are given $\alpha = \sum_{k=0}^{29} \frac{1}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}}$.Multiplying and dividing by $\sin 1^{\circ}$,we get:$\alpha = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{29} \frac{\sin((61+2k)^{\circ} - (60+2k)^{\circ})}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}}$Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have:$\alpha = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{29} (\cot(60+2k)^{\circ} - \cot(61+2k)^{\circ})$This is a telescoping sum:$\alpha = \frac{1}{\sin 1^{\circ}} [(\cot 60^{\circ} - \cot 61^{\circ}) + (\cot 62^{\circ} - \cot 63^{\circ}) + \dots + (\cot 118^{\circ} - \cot 119^{\circ})]$Since $\cot(180^{\circ} - 	heta) = -\cot 	heta$,we note that $\cot 119^{\circ} = -\cot 61^{\circ}$,$\cot 118^{\circ} = -\cot 62^{\circ}$,etc.However,the sum is simply $\frac{1}{\sin 1^{\circ}} (\cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \dots - \cot 119^{\circ})$.Given the structure,$\frac{\alpha}{\operatorname{cosec} 1^{\circ}} = \alpha \sin 1^{\circ} = \sum_{k=0}^{29} (\cot(60+2k)^{\circ} - \cot(61+2k)^{\circ}) = \cot 60^{\circ} - \cot 119^{\circ} \dots$ (Wait,the sum is $\cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \cot 63^{\circ} + \dots + \cot 118^{\circ} - \cot 119^{\circ}$).Evaluating the expression,we find $\frac{\alpha}{\operatorname{cosec} 1^{\circ}} = \frac{1}{\sqrt{3}}$.Thus,$\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}ight)^2 = (\sqrt{3})^2 = 3$.

Let $\alpha = \frac{1}{\sin 60^{\circ} \sin 61^{\circ}} + \frac{1}{\sin 62^{\circ} \sin 63^{\circ}} + \dots + \frac{1}{\sin 118^{\circ} \sin 119^{\circ}}$. Then the value of $\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2$ is $....$

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