In a $\triangle ABC$, prove that $\;$ $r r_1 \cot \left(\dfrac{A}{2}\right) = \Delta$
Circumradius $= R = \dfrac{a b c}{4 \Delta}$ $\;$ where $\Delta$ is the area of $\triangle ABC$
i.e. $\;$ $\Delta = \dfrac{a b c}{4 R}$
Inradius $= r = \dfrac{a \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)}{\cos \left(\dfrac{A}{2}\right)}$
Escribed radius $= r_1 = \dfrac{a \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)}{\cos \left(\dfrac{A}{2}\right)}$
$\begin{aligned}
LHS & = r r_1 \cot \left(\dfrac{A}{2}\right) \\\\
& = \dfrac{a \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)}{\cos \left(\dfrac{A}{2}\right)} \times \dfrac{a \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)}{\cos \left(\dfrac{A}{2}\right)} \times \cot \left(\dfrac{A}{2}\right) \\\\
& = \dfrac{a^2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{C}{2}\right)}{\cos^2 \left(\dfrac{A}{2}\right)} \times \dfrac{\cos \left(\dfrac{A}{2}\right)}{\sin \left(\dfrac{A}{2}\right)} \\\\
& \left[\text{Note: } \dfrac{\sin \theta}{2} = \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)\right] \\\\
& = a^2 \times \dfrac{\sin B}{2} \times \dfrac{\sin C}{2} \times \dfrac{2}{\sin A} \\\\
& \left[\text{Note: By sine rule, } \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R \right. \\
& \left. \implies \sin A = \dfrac{a}{2 R}, \; \sin B = \dfrac{b}{2 R}, \; \sin C = \dfrac{c}{2 R} \right] \\\\
& = a^2 \times \dfrac{1}{2} \times \dfrac{b}{2 R} \times \dfrac{c}{2 R} \times \dfrac{2 R}{a} \\\\
& = \dfrac{a b c}{4 R} \\\\
& = \Delta = RHS
\end{aligned}$
Hence proved.