In a $\triangle ABC$, prove that $\;$ $\dfrac{r \; r_1}{r_2 \; r_3} = \tan^2 \left(\dfrac{A}{2}\right)$
Inradius $= r = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)$
Escribed radius $= r_1 = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)$
Escribed radius $= r_2 = 4 R \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right)$
Escribed radius $= r_3 = 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right)$
where $\;$ $R$ $\;$ is the circumradius of $\triangle ABC$.
Now,
$\begin{aligned}
rr_1 & = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \times 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \\\\
& = 4 R^2 \sin^2 \left(\dfrac{A}{2}\right) \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) \times 2 \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{C}{2}\right) \\\\
& \left[\text{Note: }\sin \theta = 2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)\right] \\\\
& = 4 R^2 \sin^2 \left(\dfrac{A}{2}\right) \sin B \sin C \;\;\; \cdots \; (1a)
\end{aligned}$
$\begin{aligned}
r_2 r_3 & = 4 R \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \times 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \\\\
& = 4 R^2 \cos^2 \left(\dfrac{A}{2}\right) \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) \times 2 \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{C}{2}\right) \\\\
& = 4 R^2 \cos^2 \left(\dfrac{A}{2}\right) \sin B \sin C \;\;\; \cdots \; (1b)
\end{aligned}$
$\therefore \;$ We have from equations $(1a)$ and $(1b)$,
$\begin{aligned}
LHS = \dfrac{r r_1}{r_2 r_3} & = \dfrac{4 R^2 \sin^2 \left(\dfrac{A}{2}\right) \sin B \sin C}{4 R^2 \cos^2 \left(\dfrac{A}{2}\right) \sin B \sin C} \\\\
& = \dfrac{\sin^2 \left(\dfrac{A}{2}\right)}{\cos^2 \left(\dfrac{A}{2}\right)} \\\\
& = \tan^2 \left(\dfrac{A}{2}\right) = RHS
\end{aligned}$
Hence proved.