Let x= sin 1o, then the value of the expression 1cos⁡0∘⋅cos⁡1∘+1cos⁡1∘⋅cos⁡2∘+1cos⁡2∘⋅cos⁡3∘+⋯+1cos⁡44∘⋅cos⁡45∘=

1sin⁡1∘sin⁡1∘−0∘cos⁡0∘cos⁡1∘+sin⁡2∘−1∘cos⁡1∘cos⁡2∘+sin⁡3∘−2∘cos⁡2∘cos⁡3∘+⋯+sin⁡45∘−44∘cos⁡44∘cos⁡45∘=1sin⁡1∘tan⁡1∘+tan⁡2∘−tan⁡1∘+tan⁡3∘−tan⁡2∘+tan⁡4∘−tan⁡3∘+⋯+tan⁡45∘−tan⁡44∘=1sin⁡1∘=1x

Let x= sin 1o, then the value of the expression 1cos⁡0∘⋅cos⁡1∘+1cos⁡1∘⋅cos⁡2∘+1cos⁡2∘⋅cos⁡3∘+⋯+1cos⁡44∘⋅cos⁡45∘=

1sin⁡1∘sin⁡1∘−0∘cos⁡0∘cos⁡1∘+sin⁡2∘−1∘cos⁡1∘cos⁡2∘+sin⁡3∘−2∘cos⁡2∘cos⁡3∘+⋯+sin⁡45∘−44∘cos⁡44∘cos⁡45∘=1sin⁡1∘tan⁡1∘+tan⁡2∘−tan⁡1∘+tan⁡3∘−tan⁡2∘+tan⁡4∘−tan⁡3∘+⋯+tan⁡45∘−tan⁡44∘=1sin⁡1∘=1x

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Let x= sin 1^o, then the value of the expression $\frac{1}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \frac{1}{\cos 1^{\circ} \cdot \cos 2^{\circ}} + \frac{1}{\cos 2^{\circ} \cdot \cos 3^{\circ}} + \dots + \frac{1}{\cos 44^{\circ} \cdot \cos 45^{\circ}} =$

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1/x

$\frac{\sqrt{2}}{x}$

$\frac{x}{\sqrt{2}}$

answer is B.

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Detailed Solution

$\frac{1}{\sin 1^{\circ}} [\frac{\sin (1^{\circ} - 0^{\circ})}{\cos 0^{\circ} \cos 1^{\circ}} + \frac{\sin (2^{\circ} - 1^{\circ})}{\cos 1^{\circ} \cos 2^{\circ}} + \frac{\sin (3^{\circ} - 2^{\circ})}{\cos 2^{\circ} \cos 3^{\circ}} + \dots + \frac{\sin (45^{\circ} - 44^{\circ})}{\cos 44^{\circ} \cos 45^{\circ}}]$

$= \frac{1}{\sin 1^{\circ}} [\tan 1^{\circ} + (\tan 2^{\circ} - \tan 1^{\circ}) + (\tan 3^{\circ} - \tan 2^{\circ}) + (\tan 4^{\circ} - \tan 3^{\circ}) + \dots + (\tan 45^{\circ} - \tan 44^{\circ})]$