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प्रश्न
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
विकल्प
sec a1 − sec an
cosec a1 − cosec an
cot a1 − cot an
tan a1 − tan an
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उत्तर
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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