The sum 1sin45°⋅sin46°+1sin47°⋅sin48°+1sin49°⋅sin50°+⋯+1sin133°⋅sin134° is equal to

G.E. =1sin1°[sin(46°−45°)sin45°⋅sin46°+sin(48°−47°)sin47°⋅sin48°+⋯+sin(134°−133°)sin133°⋅sin134°]=1sin1°[(cot45°−cot46°)+(cot47°−cot48°)+⋯+(cot133°−cot134°)]=1sin1°[(cot45°+cot47°+⋯+cot133°)−(cot46°+cot48°+⋯+cot134°)]=cosec1° [∵ all terms cancelled except cot45°].

The sum 1sin45°⋅sin46°+1sin47°⋅sin48°+1sin49°⋅sin50°+⋯+1sin133°⋅sin134° is equal to

G.E. =1sin1°[sin(46°−45°)sin45°⋅sin46°+sin(48°−47°)sin47°⋅sin48°+⋯+sin(134°−133°)sin133°⋅sin134°]=1sin1°[(cot45°−cot46°)+(cot47°−cot48°)+⋯+(cot133°−cot134°)]=1sin1°[(cot45°+cot47°+⋯+cot133°)−(cot46°+cot48°+⋯+cot134°)]=cosec1° [∵ all terms cancelled except cot45°].

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