If frac{1}{sin 45^{circ} sin 46^{circ}}+frac{ | vedclass

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(C) The given series is $S = \sum_{k=0}^{44} \frac{1}{\sin(45^{\circ}+k) \sin(46^{\circ}+k)}$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \

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(C) The given series is $S = \sum_{k=0}^{44} \frac{1}{\sin(45^{\circ}+k) \sin(46^{\circ}+k)}$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we multiply and divide by $\sin 1^{\circ}$:$S = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{44} \frac{\sin((46^{\circ}+k) - (45^{\circ}+k))}{\sin(45^{\circ}+k) \sin(46^{\circ}+k)}$$S = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{44} (\cot(45^{\circ}+k) - \cot(46^{\circ}+k))$This is a telescoping series:$S = \frac{1}{\sin 1^{\circ}} [(\cot 45^{\circ} - \cot 46^{\circ}) + (\cot 46^{\circ} - \cot 47^{\circ}) + \ldots + (\cot 89^{\circ} - \cot 90^{\circ})]$$S = \frac{1}{\sin 1^{\circ}} [\cot 45^{\circ} - \cot 90^{\circ}]$Since $\cot 45^{\circ} = 1$ and $\cot 90^{\circ} = 0$,we have $S = \frac{1}{\sin 1^{\circ}} [1 - 0] = \frac{1}{\sin 1^{\circ}}$.Given $S = \frac{1}{\sin x^{\circ}}$,we get $x = 1$.Therefore,$\sin \left(\frac{\pi}{2} xight) = \sin \left(\frac{\pi}{2} 	imes 1ight) = \sin \left(\frac{\pi}{2}ight) = 1$.

If $\frac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{1}{\sin 46^{\circ} \sin 47^{\circ}}+\ldots$ up to $45$ terms $=\frac{1}{\sin x^{\circ}}$,then $\sin \left(\frac{\pi}{2} x\right)=$

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