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 If a1,a2,a3,.an are in A.P with common difference d0, then sum of the 

 series sindseca1seca2+seca2seca3++secan1secan is 

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a
cosecan−cos⁡eca
b
cot⁡an−cot⁡a
c
sec⁡an−sec⁡a1
d
tan⁡an−tan⁡a1

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detailed solution

Correct option is D

As a1,a2,a3,…………an−1,an are in A.P hence d=a2−a1=a3−a2=……..=an−an−1sin⁡dsec⁡a1sec⁡a2+sec⁡a2sec⁡a3+…………….+sec⁡an−1sec⁡an=sin⁡a2−a1cos⁡a1cos⁡a2+sin⁡a3−a2cos⁡a2cos⁡a3+…………..+sin⁡an−an−1cos⁡an−1cos⁡an=tan⁡a2−tan⁡a1+tan⁡a3−tan⁡a2+…………+tan⁡an−tan⁡an−1=tan⁡an−tan⁡a1


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