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 $\sum \frac{a}{s-a} = \frac{4R+r}{r} - 3 = \frac{4R}{r} + 1 - 3 = \frac{4R}{r} - 2$.

List-$I$	List-$II$
$(A) \ r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}}$	$1. \ b$
$(B) \ \frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$	$2. \ a^2, b^2, c^2 \text{ are in } AP$
$(C) \ \frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}$	$3. \ \Delta$
$(D) \ bc \cos^2 \frac{A}{2}$	$4. \ R r_1 r_2 r_3$
	$5. \ s(s-a)$

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

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