In a $\triangle ABC$, prove that $\;$ $\left(r_1 + r_2\right) \tan \left(\dfrac{C}{2}\right) = \left(r_3 - r\right) \cot \left(\dfrac{C}{2}\right) = c$
Escribed radius $= r_1 = \dfrac{\Delta}{s - a}$
Escribed radius $= r_2 = \dfrac{\Delta}{s - b}$
$\tan \left(\dfrac{C}{2}\right) = \dfrac{\Delta}{s \left(s - c\right)}$
$\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ $\;$ is the area of $\triangle ABC$
$\implies$ $\Delta^2 = s \left(s - a\right) \left(s - b\right) \left(s - c\right)$
$s = \dfrac{a + b + c}{2}$ $\;$ is the semi-perimeter of $\triangle ABC$
$\implies$ $2 s = a + b + c$
$\begin{aligned}
\therefore \; \left(r_1 + r_2\right) \tan \left(\dfrac{C}{2}\right) & = \left[\dfrac{\Delta}{s - a} + \dfrac{\Delta}{s - b}\right] \times \dfrac{\Delta}{s \left(s - c\right)} \\\\
& = \Delta^2 \times \left[\dfrac{s - b + s - a}{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}\right] \\\\
& = 2 s - a - b \\\\
& = a + b + c - a - b = c \;\;\; \cdots \; (1)
\end{aligned}$
Escribed radius $= r_3 = s \tan \left(\dfrac{C}{2}\right)$
In-radius $= r = \left(s - c\right) \tan \left(\dfrac{C}{2}\right)$
$\begin{aligned}
\therefore \; \left(r_3 - r\right) \cot \left(\dfrac{C}{2}\right) & = r_3 \cot \left(\dfrac{C}{2}\right) - r \cot \left(\dfrac{C}{2}\right) \\\\
& = s \tan \left(\dfrac{C}{2}\right) \times \cot \left(\dfrac{C}{2}\right) - \left(s - c\right) \tan \left(\dfrac{C}{2}\right) \times \cot \left(\dfrac{C}{2}\right) \\\\
& = s - \left(s - c\right) = c \;\;\; \cdots \; (2)
\end{aligned}$
$\therefore \;$ We have from equations $(1)$ and $(2)$
$\left(r_1 + r_2\right) \tan \left(\dfrac{C}{2}\right) = \left(r_3 - r\right) \cot \left(\dfrac{C}{2}\right) = c$
Hence proved.