text{Given, } frac{sin 1^{circ}}{sin x^{circ} | vedclass

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(D) We are given the identity: $\frac{\sin 1^{\circ}}{\sin x^{\circ} \sin (x+1)^{\circ}} = \cot x^{\circ} - \cot (x+1)^{\circ}$.Dividing both sides by

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$\operatorname{cosec} 1^{\circ}$

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(D) We are given the identity: $\frac{\sin 1^{\circ}}{\sin x^{\circ} \sin (x+1)^{\circ}} = \cot x^{\circ} - \cot (x+1)^{\circ}$.Dividing both sides by $\sin 1^{\circ}$,we get: $\frac{1}{\sin x^{\circ} \sin (x+1)^{\circ}} = \frac{\cot x^{\circ} - \cot (x+1)^{\circ}}{\sin 1^{\circ}}$.Let $S = \sum_{x=45}^{89} \frac{1}{\sin x^{\circ} \sin (x+1)^{\circ}}$.Substituting the identity: $S = \frac{1}{\sin 1^{\circ}} \sum_{x=45}^{89} (\cot x^{\circ} - \cot (x+1)^{\circ})$.This is a telescoping sum: $S = \frac{1}{\sin 1^{\circ}} [(\cot 45^{\circ} - \cot 46^{\circ}) + (\cot 46^{\circ} - \cot 47^{\circ}) + \dots + (\cot 89^{\circ} - \cot 90^{\circ})]$.All intermediate terms cancel out: $S = \frac{1}{\sin 1^{\circ}} (\cot 45^{\circ} - \cot 90^{\circ})$.Since $\cot 45^{\circ} = 1$ and $\cot 90^{\circ} = 0$,we have $S = \frac{1}{\sin 1^{\circ}} (1 - 0) = \frac{1}{\sin 1^{\circ}} = \operatorname{cosec} 1^{\circ}$.

$\text{Given, } \frac{\sin 1^{\circ}}{\sin x^{\circ} \sin (x+1)^{\circ}} = \cot x^{\circ} - \cot (x+1)^{\circ}, \text{ then the value of } \frac{1}{\sin 45^{\circ} \sin 46^{\circ}} + \frac{1}{\sin 46^{\circ} \sin 47^{\circ}} + \dots + \frac{1}{\sin 89^{\circ} \sin 90^{\circ}} \text{ is}$

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