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

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(D) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of $	riangle ABC$. Then $a+b+c = 2s$, $b+c-a = 2(s-a)$, $c+a-b = 2(s-b)$, and $a+b-c = 2(s-c)$.Su

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$\sin^2 A$

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(D) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of $	riangle ABC$. Then $a+b+c = 2s$, $b+c-a = 2(s-a)$, $c+a-b = 2(s-b)$, and $a+b-c = 2(s-c)$.Substituting these into the expression:$\frac{(2s)(2(s-a))(2(s-b))(2(s-c))}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2}$.Using Heron's formula, $\Delta^2 = s(s-a)(s-b)(s-c)$, so the expression becomes $\frac{16\Delta^2}{4b^2c^2} = \frac{4\Delta^2}{b^2c^2}$.Since the area $\Delta = \frac{1}{2}bc \sin A$, we have $\sin A = \frac{2\Delta}{bc}$.Thus, $\frac{4\Delta^2}{b^2c^2} = (\frac{2\Delta}{bc})^2 = \sin^2 A$.

In any $\triangle ABC$, 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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