I. Since, $4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$$
\begin{aligned}
& =\frac{4 a b c}{4 \Delta} \sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{(s-a)(s-c)}{a c}} \\
& =\frac{a b c}{\Delta} \sqrt{\frac{(s-a)(s-b)}{a b}} \\
& =\frac{a b c}{4 \Delta} \frac{\Delta^2}{s \cdot a b c} \\
& =\frac{\Delta}{s}=r
\end{aligned}
$$
Hence, $r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
II. Since,
$$
\begin{aligned}
& (s-a) \tan A / 2=(s-a) \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \\
& =\sqrt{\frac{s(s-a)(s-b)(s-c)}{s^2}} \\
& =\frac{\Delta}{s}=r
\end{aligned}
$$
Hence, $r_1 \neq(s-a) \tan \frac{A}{2}$
III. Since, $\frac{\Delta}{s-c}=\sqrt{\frac{s(s-a)(s-b)}{(s-c)}}$
$$
=s \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=s \tan \frac{c}{2}=r_3
$$
Hence $r_3=\frac{\Delta}{s-c}$
Therefore I and III statements are true but II is a false statement.