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In triangleABC, s(s-a)tan((A)/(2))=...

In `triangleABC, s(s-a)tan((A)/(2))=`

A

`(A(triangleABC))/(2)`

B

`(2A(triangleABC))/(4)`

C

`A(triangleABC)`

D

`2A(triangleABC)`

Text Solution

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The correct Answer is:
C
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