Question

In a $△ABC,\frac{r_2-r}{r_1+r_3}=$

• A. ${\sin }^2\frac{A}{2}$
• B. ${\sin }^2\frac{B}{2}$
• C. ${\tan }^2\frac{A}{2}$
• D. ${\tan }^2\frac{B}{2}$

Answer 1

The correct option is D ${\tan }^2\frac{B}{2}$

We know $r_2-r=\frac{\Delta }{s-b}-\frac{\Delta }{s}=\frac{b\Delta }{s(s-b)}$

and $r_1+r_3=\frac{\Delta }{s-a}+\frac{\Delta }{s-c}=\frac{\Delta (2s-a-c)}{(s-a)(s-c)}=\frac{b\Delta }{(s-a)(s-c)}$

Therefore, $\frac{r_2-r}{r_1+r_3}=\frac{b\Delta }{s(s-b)}\cdot \frac{(s-a)(s-c)}{b\Delta }$

$=\frac{(s-a)(s-c)}{s(s-b)}={\tan }^2\frac{B}{2}$