In any triangle ABC,a cdot cos^2 frac{A}{2} + | vedclass

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(B) We know that $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$.Substituting this into the expression,we get:$a \cdot \frac{s(s-a)}{bc} + b \cdot \frac{s(s-

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$s + \frac{\Delta}{R}$

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(B) We know that $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$.Substituting this into the expression,we get:$a \cdot \frac{s(s-a)}{bc} + b \cdot \frac{s(s-b)}{ac} + c \cdot \frac{s(s-c)}{ab}$$= \frac{s}{abc} [a^2(s-a) + b^2(s-b) + c^2(s-c)]$$= \frac{s}{abc} [s(a^2+b^2+c^2) - (a^3+b^3+c^3)]$Using the identity $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$ and $s = \frac{a+b+c}{2}$,we simplify the expression to $s + \frac{\Delta}{R}$.

In any triangle $ABC$,$a \cdot \cos^2 \frac{A}{2} + b \cdot \cos^2 \frac{B}{2} + c \cdot \cos^2 \frac{C}{2} =$

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