With usual notations, in triangle ABC, (b - c | vedclass

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(A) Given expression: $E = (b - c)^2 \cos^2 \frac{A}{2} + (b + c)^2 \sin^2 \frac{A}{2}$Using half-angle formulas: $\cos^2 \frac{A}{2} = \frac{s(s-a)}{

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$a^2$

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(A) Given expression: $E = (b - c)^2 \cos^2 \frac{A}{2} + (b + c)^2 \sin^2 \frac{A}{2}$Using half-angle formulas: $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$ and $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$, where $s = \frac{a+b+c}{2}$.$E = (b-c)^2 \frac{s(s-a)}{bc} + (b+c)^2 \frac{(s-b)(s-c)}{bc}$$E = \frac{1}{bc} [ (b^2 - 2bc + c^2)s(s-a) + (b^2 + 2bc + c^2)(s-b)(s-c) ]$Since $s-b = \frac{a-b+c}{2}$ and $s-c = \frac{a+b-c}{2}$, $(s-b)(s-c) = \frac{a^2 - (b-c)^2}{4}$ and $s(s-a) = \frac{(b+c)^2 - a^2}{4}$.Substituting these, the expression simplifies to $a^2$.

With usual notations, in $\triangle ABC$, $(b - c)^2 \cos^2 \frac{A}{2} + (b + c)^2 \sin^2 \frac{A}{2} = $

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