The sum $\frac{1}{\sin 45 ° \cdot \sin 46 °} + \frac{1}{\sin 47 ° \cdot \sin 48 °} + \frac{1}{\sin 49 ° \cdot \sin 50 °} + \dots$ $+ \frac{1}{\sin 133 ° \cdot \sin 134 °}$ is equal to

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$\tan 1 °$

$\sec 1 °$

$\cot 1 °$

$\cos e c 1 °$

answer is B.

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Detailed Solution

G.E. $= \frac{1}{\sin 1 °} [\frac{\sin (46 ° - 45 °)}{\sin 45 ° \cdot \sin 46 °} + \frac{\sin (48 ° - 47 °)}{\sin 47 ° \cdot \sin 48 °} + \dots + \frac{\sin (134 ° - 133 °)}{\sin 133 ° \cdot \sin 134 °}]$

$= \frac{1}{\sin 1 °} [(\cot 45 ° - \cot 46 °) + (\cot 47 ° - \cot 48 °) + \dots + (\cot 133 ° - \cot 134 °)]$

$= \frac{1}{\sin 1 °} [(\cot 45 ° + \cot 47 ° + \dots + \cot 133 °) - (\cot 46 ° + \cot 48 ° + \dots + \cot 134 °)]$

$= \cos e c 1 °$

[ $∵$ all terms cancelled except $\cot 45 °$ ].

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