If a1,a2,…,an are in AP with common difference d, then the sum of the series sind cosec ⁡a1cosec ⁡a2+cosec ⁡a2cosec ⁡a3+…+cosec⁡an−1cosec⁡an is

A
$\sec a_{1} - \sec a_{n}$
B
$\cot a_{1} - \cot a_{n}$
C
$\tan a_{1} - \tan a_{n}$
D
$cosec a_{1} - cosec a_{n}$

Solution:

Since, $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in AP

$\Rightarrow d = a_{2} - a_{1} = a_{3} - a_{2} = \dots = a_{n} - a_{n^{'} - 1}$

$\begin{array}{l} ∴ \sin d (cosec a_{1} cosec a_{2} + \dots + cosec a_{n - 1} cosec a_{n}) \\ = \frac{\sin (a_{2} - a_{1})}{\sin a_{1} \sin a_{2}} + \dots + \frac{\sin (a_{n} - a_{n - 1})}{\sin a_{n - 1} \sin a_{n}} \\ = \frac{(\sin a_{2} \cos a_{1} - \cos a_{2} \sin a_{1})}{\sin a_{1} \sin a_{2}} + \dots + \frac{(\sin a_{n} \cos a_{n - 1} + \cos a_{n} \sin a_{n - 1})}{\sin a_{n - 1} \sin a_{n}} \end{array}$

$\begin{aligned} = (\cot a_{1} - \cot a_{2}) + (\cot a_{2} - \cot a_{3}) \end{aligned}$ $+ \dots + (\cot a_{n - 1} - \cot a_{n})$

$= \cot a_{1} - \cot a_{n}$

If $a_{1}, a_{2}, \dots, a_{n}$ are in AP with common difference d, then the sum of the series sind $(cosec a_{1} cosec a_{2} + cosec a_{2} cosec a_{3} + \dots +$ $cosec a_{n - 1} cosec a_{n})$ is

Solution:

Related content