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

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(C) Let the given sum be $S = \sum_{k=1}^{89} \frac{1}{\sin k^{\circ} \sin(k+1)^{\circ}}$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the given sum be $S = \sum_{k=1}^{89} \frac{1}{\sin k^{\circ} \sin(k+1)^{\circ}}$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we can write $1 = \sin((k+1)^{\circ} - k^{\circ}) = \sin(k+1)^{\circ} \cos k^{\circ} - \cos(k+1)^{\circ} \sin k^{\circ}$.Dividing by $\sin k^{\circ} \sin(k+1)^{\circ}$,we get $\frac{1}{\sin k^{\circ} \sin(k+1)^{\circ}} = \frac{\sin(k+1)^{\circ} \cos k^{\circ} - \cos(k+1)^{\circ} \sin k^{\circ}}{\sin k^{\circ} \sin(k+1)^{\circ}} = \cot k^{\circ} - \cot(k+1)^{\circ}$.Thus,$S = \sum_{k=1}^{89} (\cot k^{\circ} - \cot(k+1)^{\circ}) = (\cot 1^{\circ} - \cot 2^{\circ}) + (\cot 2^{\circ} - \cot 3^{\circ}) + \ldots + (\cot 89^{\circ} - \cot 90^{\circ})$.This is a telescoping sum,so $S = \cot 1^{\circ} - \cot 90^{\circ}$.Since $\cot 90^{\circ} = 0$,we have $S = \cot 1^{\circ} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$.Comparing with the options,$\frac{\cot 1^{\circ}}{\sin 1^{\circ}}$ is not correct,but $\frac{\cos 1^{\circ}}{\sin 1^{\circ}}$ is $\cot 1^{\circ}$. Note that $\frac{\cos 1^{\circ}}{\sin 1^{\circ}} = \frac{\cos^2 1^{\circ}}{\sin 1^{\circ} \cos 1^{\circ}}$. Actually,checking the options again,none match $\cot 1^{\circ}$ directly. However,if we evaluate $\frac{\cot 1^{\circ}}{\sin 1^{\circ}}$ it is $\frac{\cos 1^{\circ}}{\sin^2 1^{\circ}}$. Let us re-verify. The sum is $\frac{1}{\sin 1^{\circ}} \sum (\cot k^{\circ} - \cot(k+1)^{\circ}) 	imes \sin 1^{\circ}$. The correct value is $\frac{\cot 1^{\circ}}{\sin 1^{\circ}}$ is incorrect. The correct answer is $\frac{\cot 1^{\circ}}{\sin 1^{\circ}}$ is not matching. Wait,$\frac{1}{\sin 1^{\circ}} \sum \frac{\sin(1^{\circ})}{\sin k^{\circ} \sin(k+1)^{\circ}} = \frac{1}{\sin 1^{\circ}} (\cot 1^{\circ} - \cot 90^{\circ}) = \frac{\cot 1^{\circ}}{\sin 1^{\circ}}$. Thus option $C$ is correct.

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

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