The integer part of the number sum_{k=0}^{44} | vedclass

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(C) Let $S = \sum_{k=0}^{44} \frac{1}{\cos k^{\circ} \cos (k+1)^{\circ}}$.Multiply and divide by $\sin 1^{\circ}$:$S = \frac{1}{\sin 1^{\circ}} \sum_{

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(C) Let $S = \sum_{k=0}^{44} \frac{1}{\cos k^{\circ} \cos (k+1)^{\circ}}$.Multiply and divide by $\sin 1^{\circ}$:$S = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{44} \frac{\sin((k+1)^{\circ} - k^{\circ})}{\cos k^{\circ} \cos (k+1)^{\circ}}$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:$S = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{44} (	an (k+1)^{\circ} - 	an k^{\circ})$.This is a telescoping sum:$S = \frac{1}{\sin 1^{\circ}} [(	an 1^{\circ} - 	an 0^{\circ}) + (	an 2^{\circ} - 	an 1^{\circ}) + \dots + (	an 45^{\circ} - 	an 44^{\circ})]$.$S = \frac{1}{\sin 1^{\circ}} (	an 45^{\circ} - 	an 0^{\circ}) = \frac{1}{\sin 1^{\circ}} (1 - 0) = \frac{1}{\sin 1^{\circ}}$.Since $\sin 1^{\circ} \approx 0.01745$,we have $S \approx \frac{1}{0.01745} \approx 57.299$.The integer part of $57.299$ is $57$.

The integer part of the number $\sum_{k=0}^{44} \frac{1}{\cos k^{\circ} \cos (k+1)^{\circ}}$ is

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