In a triangle ABC,the expression frac{(a+b+c) | vedclass

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(C) Let $2s = a+b+c$. Then $b+c-a = 2s-2a$,$c+a-b = 2s-2b$,and $a+b-c = 2s-2c$. Substituting these into the expression: $\frac{(2s)(2s-2a)(2s-2b)(2s-2

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$\sin^2 A$

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(C) Let $2s = a+b+c$. Then $b+c-a = 2s-2a$,$c+a-b = 2s-2b$,and $a+b-c = 2s-2c$. Substituting these into the expression: $\frac{(2s)(2s-2a)(2s-2b)(2s-2c)}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2} = 4 \frac{s(s-a)}{bc} \cdot \frac{(s-b)(s-c)}{bc}$. Using the half-angle formulas $\cos^2(\frac{A}{2}) = \frac{s(s-a)}{bc}$ and $\sin^2(\frac{A}{2}) = \frac{(s-b)(s-c)}{bc}$,we get: $4 \cos^2(\frac{A}{2}) \sin^2(\frac{A}{2}) = (2 \sin(\frac{A}{2}) \cos(\frac{A}{2}))^2 = \sin^2 A$.

In a $\triangle ABC$,the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

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