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If A1, A2, A3, .... an Are in A.P. with Common Difference D, Then the Sum of the Series Sin D [Cosec A1 Cosec A2 + Cosec A1 Cosec A3 + .... + Cosec an − 1 Cosec An] is

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Question

If a1, a2, a3, .... an are in A.P. with common difference d, then the sum of the series sin d [cosec a1cosec a2 + cosec a1 cosec a3 + .... + cosec an − 1 cosec an] is

Options

  •  sec a1 − sec an

  • cosec a1 − cosec an

  • cot a1 − cot an

  • tan a1 − tan an

MCQ
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Solution

 cot a1 − cot an

We have:

\[\sin d \left( \cos ec \ a_1 \cos ec \ a_2 + cos \ ec a_2 \cos ec \ a_3 + . . . . + \cos ec a_{n - 1} \cos ec \ a_n  \right)\]

\[ = \frac{\sin d}{\sin a_1 \sin a_2} + \frac{\sin d}{\sin a_2 \sin a_3} + . . . . . + \frac{\sin d}{\sin a_{n - 1} \sin a_n}\]

\[ = \frac{\sin ( a_2 - a_1 )}{\sin a_1 \sin a_2} + \frac{\sin ( a_3 - a_2 )}{\sin a_2 \sin a_3} + . . . . + \frac{\sin ( a_n - a_{n - 1} )}{\sin a_{n - 1} \sin a_n}\]

\[ = \frac{\sin a_2 \cos a_1 - \cos a_2 \sin a_1}{\sin a_1 \sin a_2} + \frac{\sin a_3 \cos a_2 - \cos a_3 \sin a_2}{\sin a_1 \sin a_2} + . . . . . + \frac{\sin a_2 \cos a_1 - \cos a_2 \sin a_1}{\sin a_1 \sin a_2}\]

\[ = \left( \cot a_1 - \cot a_2 \right) + \left( \cot a_2 - \cot a_3 \right) + . . . . . + \left( \cot a_{n - 1} - \cot a_n \right)\]

\[ = \cot a_1 - \cot a_n\]

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