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

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Question

(C) माना योग $S = \sum_{k=1}^{89} \frac{1}{\sin k^{\circ} \sin(k+1)^{\circ}}$ है।$\sin(1^{\circ}) = \sin((k+1)^{\circ} - k^{\circ}) = \sin(k+1)^{\circ

Vedclass · Accepted Answer

$\frac{\cot 1^{\circ}}{\sin 1^{\circ}}$

Vedclass · Answer

(C) माना योग $S = \sum_{k=1}^{89} \frac{1}{\sin k^{\circ} \sin(k+1)^{\circ}}$ है।$\sin(1^{\circ}) = \sin((k+1)^{\circ} - k^{\circ}) = \sin(k+1)^{\circ} \cos k^{\circ} - \cos(k+1)^{\circ} \sin k^{\circ}$ का उपयोग करते हुए।अतः,$\frac{1}{\sin k^{\circ} \sin(k+1)^{\circ}} = \frac{1}{\sin 1^{\circ}} (\cot k^{\circ} - \cot(k+1)^{\circ})$।इसलिए,$S = \frac{1}{\sin 1^{\circ}} (\cot 1^{\circ} - \cot 90^{\circ}) = \frac{\cot 1^{\circ}}{\sin 1^{\circ}}$।

$\frac{1}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{1}{\sin 2^{\circ} \sin 3^{\circ}}+\frac{1}{\sin 3^{\circ} \sin 4^{\circ}}+\ldots+\frac{1}{\sin 89^{\circ} \sin 90^{\circ}} = ?$

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