frac{1}{sin 1^{circ} sin 2^{circ}}+frac{1}{si | vedclass

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Question

(B) Let the given expression be $S$. $S = \sum_{k=1}^{89} \frac{1}{\sin k^{\circ} \sin(k+1)^{\circ}}$ Multiply and divide by $\sin 1^{\circ}$: $S = \f

Vedclass · Accepted Answer

$\frac{\cos 1^{\circ}}{\sin ^2 1^{\circ}}$

Vedclass · Answer

(B) Let the given expression be $S$. $S = \sum_{k=1}^{89} \frac{1}{\sin k^{\circ} \sin(k+1)^{\circ}}$ Multiply and divide by $\sin 1^{\circ}$: $S = \frac{1}{\sin 1^{\circ}} \sum_{k=1}^{89} \frac{\sin((k+1)^{\circ} - k^{\circ})}{\sin k^{\circ} \sin(k+1)^{\circ}}$ Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $S = \frac{1}{\sin 1^{\circ}} \sum_{k=1}^{89} \left( \frac{\sin(k+1)^{\circ} \cos k^{\circ} - \cos(k+1)^{\circ} \sin k^{\circ}}{\sin k^{\circ} \sin(k+1)^{\circ}} \right)$ $S = \frac{1}{\sin 1^{\circ}} \sum_{k=1}^{89} (\cot k^{\circ} - \cot(k+1)^{\circ})$ This is a telescoping sum: $S = \frac{1}{\sin 1^{\circ}} [(\cot 1^{\circ} - \cot 2^{\circ}) + (\cot 2^{\circ} - \cot 3^{\circ}) + \ldots + (\cot 89^{\circ} - \cot 90^{\circ})]$ $S = \frac{1}{\sin 1^{\circ}} (\cot 1^{\circ} - \cot 90^{\circ})$ Since $\cot 90^{\circ} = 0$: $S = \frac{\cot 1^{\circ}}{\sin 1^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ} \cdot \sin 1^{\circ}} = \frac{\cos 1^{\circ}}{\sin^2 1^{\circ}}$

$\frac{1}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{1}{\sin 2^{\circ} \sin 3^{\circ}}+\ldots+\frac{1}{\sin 89^{\circ} \sin 90^{\circ}} = $

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