If $a_{1}, a_{2}, . . . ., a_{n}$ are in $A P$ with common difference $d$ , then the sum of the series $sin d (csc a_{1} csc a_{2} + csc a_{2} csc a_{3} + . . . + csc a_{n - 1} csc a_{n})$ is

sec a_{1} - sec a_{n}

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cot a_{1} - cot a_{n}

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tan a_{1} - tan a_{n}

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csc a_{1} - csc a_{n}

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Solution

The correct option is B $cot a_{1} - cot a_{n}$
Given
$a_{1}, a_{2}, a_{3} . . . . . a n$ be A.P with d as common difference, then
$d = a_{2} - a_{1} = a_{3} - a_{2} . . . . . . . . . = a_{n} - a_{n - 1}$
$sin d [csc a_{1} csc a_{2} + csc a_{2} csc 2_{3} + . . . . + csc a_{n - 1} csc a_{n}]$
$= \frac{sin d}{sin a_{1} sin a_{2}} + \frac{sin d}{sin a_{2} sin a_{3}} . . . . + \frac{sin d}{sin a_{n - 1} sin a_{n}}$

$= \frac{sin (a_{2} - a_{1})}{sin a_{1} sin a_{2}} + \frac{sin (a_{3} - a_{2})}{sin a_{2} sin a_{3}} . . . . + \frac{sin (a_{n} - a_{n - 1})}{sin a_{n - 1} sin a_{n}}$

$= \frac{sin a_{1} cos a_{2} - cos a_{1} sin a_{2}}{sin a_{1} sin a_{2}} + \frac{sin a_{2} cos a_{3} - cos a_{2} sin a_{3}}{sin a_{2} sin a_{3}} . . . . + \frac{sin a_{n} cos a_{n - 1} - cos a_{n - 1} sin a_{n}}{sin a_{n - 1} sin a_{n}}$

$= cot a_{1} - cot a_{2} + cot a_{2} - cot a_{3} + . . . . . + cot a_{n - 1} - cot a_{n}$
$= cot a_{1} - cot a_{n}$

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