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 the triangle,so $a+b+c = 2s$.Then,$b+c-a = 2s-2a$,$c+a-b = 2s-2b$,and $a+b-c = 2s-2c$.The expre

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$\sin^2 A$

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(D) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of the triangle,so $a+b+c = 2s$.Then,$b+c-a = 2s-2a$,$c+a-b = 2s-2b$,and $a+b-c = 2s-2c$.The expression becomes:$\frac{(2s)(2s-2a)(2s-2b)(2s-2c)}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2}$.Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c)$,so the expression is $\frac{16\Delta^2}{4b^2c^2} = \frac{4\Delta^2}{b^2c^2}$.This can be written as $\left(\frac{2\Delta}{bc}\right)^2$.Since the area of a triangle is $\Delta = \frac{1}{2}bc \sin A$,we have $\sin A = \frac{2\Delta}{bc}$.Therefore,the expression equals $\sin^2 A$.

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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