In a triangle ABC,sleft[frac{r_1-r}{a}+frac{r | vedclass

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(B) We know that $r = (s-a) 	an(A/2) = (s-b) 	an(B/2) = (s-c) 	an(C/2)$ and $r_1 = s 	an(A/2)$,$r_2 = s 	an(B/2)$,$r_3 = s 	an(C/2)$.Then,$r_1 -

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$r_1+r_2+r_3$

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(B) We know that $r = (s-a) 	an(A/2) = (s-b) 	an(B/2) = (s-c) 	an(C/2)$ and $r_1 = s 	an(A/2)$,$r_2 = s 	an(B/2)$,$r_3 = s 	an(C/2)$.Then,$r_1 - r = s 	an(A/2) - (s-a) 	an(A/2) = a 	an(A/2)$.Thus,$\frac{r_1-r}{a} = 	an(A/2)$.Similarly,$\frac{r_2-r}{b} = 	an(B/2)$ and $\frac{r_3-r}{c} = 	an(C/2)$.Substituting these into the expression,we get $s[	an(A/2) + 	an(B/2) + 	an(C/2)]$.Since $r_1 = s 	an(A/2)$,$r_2 = s 	an(B/2)$,and $r_3 = s 	an(C/2)$,the expression becomes $r_1 + r_2 + r_3$.

In a triangle $ABC$,$s\left[\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c}\right]=$

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