In a triangle ABC,frac{2(r_1+r_3)}{ac(1+cos B | vedclass

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(B) We know that $r_1 = \frac{\Delta}{s-a}$ and $r_3 = \frac{\Delta}{s-c}$.So,$r_1 + r_3 = \Delta \left( \frac{1}{s-a} + \frac{1}{s-c} \right) = \Delt

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$\frac{b}{\Delta}$

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(B) We know that $r_1 = \frac{\Delta}{s-a}$ and $r_3 = \frac{\Delta}{s-c}$.So,$r_1 + r_3 = \Delta \left( \frac{1}{s-a} + \frac{1}{s-c} \right) = \Delta \left( \frac{s-c+s-a}{(s-a)(s-c)} \right) = \Delta \left( \frac{2s-a-c}{(s-a)(s-c)} \right)$.Since $2s = a+b+c$,we have $2s-a-c = b$.Thus,$r_1 + r_3 = \frac{\Delta b}{(s-a)(s-c)}$.Also,$1 + \cos B = 1 + \frac{a^2+c^2-b^2}{2ac} = \frac{2ac+a^2+c^2-b^2}{2ac} = \frac{(a+c)^2-b^2}{2ac} = \frac{(a+c-b)(a+c+b)}{2ac} = \frac{(2s-2b)(2s)}{2ac} = \frac{2(s-b)s}{ac}$.Substituting these into the expression:$\frac{2(r_1+r_3)}{ac(1+\cos B)} = \frac{2 \cdot \frac{\Delta b}{(s-a)(s-c)}}{ac \cdot \frac{2s(s-b)}{ac}} = \frac{2 \Delta b}{2s(s-a)(s-b)(s-c)} = \frac{\Delta b}{s(s-a)(s-b)(s-c)}$.Since $\Delta^2 = s(s-a)(s-b)(s-c)$,the expression becomes $\frac{\Delta b}{\Delta^2} = \frac{b}{\Delta}$.

In a $\triangle ABC$,$\frac{2(r_1+r_3)}{ac(1+\cos B)} = $

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