In $ riangle \mathrm{ABC}$, if $\mathrm{r}$ is the inradius and $\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3$ are the ex-radii, then $\frac{1}{4}\left[b^2 \sin 2 C+c^2 \sin 2 B ight]=$

Question

In $	riangle \mathrm{ABC}$, if $\mathrm{r}$ is the inradius and $\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3$ are the ex-radii, then $\frac{1}{4}\left[b^2 \sin 2 C+c^2 \sin 2 Bight]=$

Aicharya · Accepted Answer

$\frac{1}{4}\left[b^2 \sin 2 C+c^2 \sin 2 Bight]$ $\begin{aligned} & =\frac{1}{4}\left[b^2 \cdot 2 \sin C \cdot \cos C+c^2 \cdot 2 \sin B \cdot \cos Bight] \ & =\frac{1}{2}\left[b^2 \cos C \cdot \frac{2 \Delta}{a b}+c^2 \cdot \cos B \cdot \frac{2 \Delta}{c a}ight]\left[\because \sin A=\frac{2 \Delta}{b c}ight] \ & =\frac{\Delta}{a}[b \cos C+c \cos B]=\Delta[\because a=b \cos C+c \cos B] \ & \because r r_1 \cot \frac{A}{2}=\frac{\Delta}{s} \cdot \frac{\Delta}{s-a} \cdot \frac{s(s-a)}{\Delta}=\Delta \ & 	herefore \frac{1}{4}\left[b^2 \sin 2 C+c^2 \sin 2 Bight]=r r_1 \cot \frac{A}{2}\end{aligned}$

In $\triangle \mathrm{ABC}$, if $\mathrm{r}$ is the inradius and $\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3$…

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