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

In a triangle $\mathrm{ABC} s\left[\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c}\right]=$
  1. $\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$
  2. $r_1+r_2+r_3$
  3. $r_1 r_2 r_3$
  4. $\frac{1}{r}-\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$

Solution

$r=(\mathrm{s}-\mathrm{a}) \tan \mathrm{A} / 2=(\mathrm{s}-\mathrm{b}) \tan \mathrm{B} / 2=(\mathrm{s}-\mathrm{c}) \tan \mathrm{C} / 2$ $\begin{aligned} & r_1=\mathrm{s} \tan \mathrm{A} / 2, r_2=\mathrm{s} \tan \mathrm{B} / 2, r_3=\mathrm{s} \tan \mathrm{C} / 2 \\ & r_1-r=\mathrm{s} \tan \mathrm{A} / 2-(\mathrm{s}-\mathrm{a}) \tan \mathrm{A} / 2=\mathrm{a} \tan \mathrm{A} / 2 \\ & \frac{r_1-r}{a}=\tan \frac{A}{2}\end{aligned}$ Similarly, $\frac{r_2-r}{b}=\tan \frac{B}{2}, \frac{r_3-r}{c}=\tan \frac{C}{2}$ $\begin{aligned} & \therefore s\left[\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\right] \\ & =s \tan \frac{A}{2}+s \tan \frac{B}{2}+s \tan \frac{C}{2} \\ & =r_1+r_2+r_3\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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