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in a triangle prove that bcos^2C/2+ccos^2B/2=s here s is semiperimeter of a triangle
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Is it (bcos2c)/2
U know half angle theorems right Use those equations to find the solution I got it by that method
As cos C/2=square root of (s(s-c))/ab. Cos B/2=square root of (s(s-b))/ac
From numerator and denominator b and c get cancelled
Then take s/a com on to get s/a(s-b+s-c) =s/a(2s-b-c) = (s/a)*a = s hence proved
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thank u i did some mistake in the last step
Ur welcome
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