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In triangleABC, sin((B-C)/(2))=...

In `triangleABC, sin((B-C)/(2))=`

A

`2((b-c)/(a))cos((A)/(2))`

B

`((b-c)/(a))cos((A)/(2))`

C

`2((c-b)/(a))cos((A)/(2))`

D

`((c-b)/(a))cos((A)/(2))`

Text Solution

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The correct Answer is:
B
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