In triangle ABC,frac{a}{s-a}+frac{b}{s-b}+fra | vedclass

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(D) We know that $s = \frac{a+b+c}{2}$,so $2s = a+b+c$. Let $E = \frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}$. Substituting $a = 2s - (b+c)$,$b = 2s - (

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$\frac{4 R}{r}-2$

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(D) We know that $s = \frac{a+b+c}{2}$,so $2s = a+b+c$. Let $E = \frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}$. Substituting $a = 2s - (b+c)$,$b = 2s - (a+c)$,$c = 2s - (a+b)$ is complex. Instead,use the identity $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s} = \Delta r$. Also,$abc = 4R\Delta$. The expression simplifies as: $\frac{a}{s-a} + \frac{b}{s-b} + \frac{c}{s-c} = \frac{a(s-b)(s-c) + b(s-a)(s-c) + c(s-a)(s-b)}{(s-a)(s-b)(s-c)}$ Using the identity $\sum a(s-b)(s-c) = 4R\Delta - 2\Delta r$ is not direct,but evaluating the sum: $\sum \frac{a}{s-a} = \sum \frac{a+s-s}{s-a} = \sum (\frac{s}{s-a} - 1) = s \sum \frac{1}{s-a} - 3$. Since $\sum \frac{1}{s-a} = \frac{1}{r}$,the expression becomes $s(\frac{1}{r}) - 3 = \frac{s}{r} - 3$. However,using the standard identity $\sum \frac{a}{s-a} = \frac{4R+r}{r} - 3 = \frac{4R}{r} + 1 - 3 = \frac{4R}{r} - 2$.

In $\triangle ABC$,$\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c} =$

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