If $a_{1}, a_{2}, . . ., a_{n},$ are in AP with a common difference $d \neq 0$ , then the sum of the following series is $(\sin d) [\cos e c a_{1} \cos e c a_{2} + \cos e c a_{2} \cos e c a_{3} + . . . + \cos e c a_{(n - 1)} \cos e c a_{n}]$ is equal to

$s e c a_{1} - s e c a_{n}$

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$c o t a_{1} - c o t a_{n}$

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$\tan a_{1} - \tan a_{n}$

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$\cos e c a_{1} - \cos e c a_{n}$

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Solution

The correct option is B
$c o t a_{1} - c o t a_{n}$

Explanation for the correct option:
Find the value of $(s i n d) [c o s e c a_{1} c o s e c a_{2} + c o s e c a_{2} c o s e c a_{3} + . . . + c o s e c a_{(n - 1)} c o s e c a_{n}]$ :
Given $a_{1}, a_{2}, \dots . . ., a_{n}$ are in AP.
Common difference, $d = a_{2} - a_{1} = a_{3} - a_{2} = a_{n} - a_{(n - 1)}$
$\begin{array}{rcl} (\sin d) [\cos e c a_{1} \cos e c a_{2} + \cos e c a_{2} \cos e c a_{3} + \dots + \cos e c a_{(n - 1)} \cos e c a_{n}] \\ = & \frac{\sin (a_{2} - a_{1})}{\sin a_{1} \sin a_{2}} + \frac{\sin (a_{3} - a_{2})}{\sin a_{2} \sin a_{3}} + \dots . . + \frac{\sin (a_{n} - a_{(n - 1)})}{\sin a_{(n - 1)} \sin a_{n}} \\ = & \frac{(\sin a_{2} \cos a_{1} - \sin a_{1} \cos a_{2})}{\sin a_{1} \sin a_{2}} + \frac{(\sin a_{3} \cos a_{2} - \sin a_{2} \cos a_{3})}{s in a_{2} \sin a_{3}} + \dots . . + \frac{(\sin a_{n} \cos a_{(n - 1)} - \sin a_{(n - 1)} \cos a_{n})}{\sin a_{(n - 1)} \sin a_{n}} [∵ s i n (A - B) = s i n A c o s B - c o s A s i n B] \\ = & (c o t a_{1} - c o t a_{2}) + (c o t a_{2} - c o t a_{3}) + \dots . + (c o t a_{n - 1} - c o t a_{n}) \\ = & (c o t a_{1} - c o t a_{n}) \end{array}$
Hence, the correct option is B.

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Q. If

a_{1}, a_{2}, . . . ., a_{n}

are in

A P

with common difference

d

, then the sum of the series

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