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Question
If a1, a2, ..., an are in A.P. with common difference d (where d ≠ 0); then the sum of the series sin d (cosec a1 cosec a2 + cosec a2 cosec a3 + ...+ cosec an–1 cosec an) is equal to cot a1 – cot an
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Solution
We have sin d (cosec a1 cosec a2 + cosec a2 cosec a3 + ...+ cosec an–1 cosec an)
= `sin d[1/(sina_1 sina_2) + 1/(sina_2 sina_3) + ... + 1/(sina_(n - 1) sina_n)]`
= `(sin(a_2 - a_1))/(sina_1 sina_2) + (sin(a_3 - a_2))/(sina_2 sina_3) + ... + (sin(a_n - a_(n - 1)))/(sina_(n - 1) sina_n)`
= `(sina_2 cos a_1 - cosa_2 sina_1)/(sina_1 sina_2) + (sina_3 cosa_2 - cosa_3 sina_2)/(sina_2 sina_3) + ... + (sina_n cosa_(n - 1) - cosa_n sina_(n - 1))/(sina_(n - 1) sina_n)`
= (cot a1 – cot a2) + (cot a2 – cot a3) + ... + (cot an–1 – cot an)
= cot a1 – cot an
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