In any $\triangle ABC$, prove that $\;$ $\left(b + c - a\right)\left[\cot\left(\dfrac{B}{2}\right) + \cot \left(\dfrac{C}{2}\right)\right] = 2 a \cot \left(\dfrac{A}{2}\right)$
In $\triangle ABC$,
$\cot \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{s \left(s - a\right)}{\left(s - b\right) \left(s - c\right)}}$ $\;\; \cdots \; (1a)$,
$\cot \left(\dfrac{B}{2}\right) = \sqrt{\dfrac{s \left(s - b\right)}{\left(s - c\right) \left(s - a\right)}}$ $\;\; \cdots \; (1b)$,
$\cot \left(\dfrac{C}{2}\right) = \sqrt{\dfrac{s \left(s - c\right)}{\left(s - a\right) \left(s - b\right)}}$ $\;\; \cdots \; (1c)$
$s = \dfrac{a + b + c}{2} = $ semi perimeter of $\triangle ABC$
i.e. $\;$ $2 s = a + b + c$ $\;\;\; \cdots \; (2a)$
and $\;$ $2 s - 2a = b + c - a$ $\;$ i.e. $\;$ $2 \left(s - a\right) = b + c - a$ $\;\;\; \cdots \; (2b)$
$\begin{aligned}
LHS & = \left(b + c - a\right)\left[\cot\left(\dfrac{B}{2}\right) + \cot \left(\dfrac{C}{2}\right)\right] \\\\
& = 2 \left(s - a\right) \left[\sqrt{\dfrac{s \left(s - b\right)}{\left(s - c\right) \left(s - a\right)}} + \sqrt{\dfrac{s \left(s - c\right)}{\left(s - a\right) \left(s - b\right)}}\right] \\\\
& \hspace{3cm} \left[\text{by equations (1b), (1c) and (2b)}\right] \\\\
& = 2 \left(s - a\right) \left[\sqrt{\dfrac{s \left(s - b\right)^2}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}} + \sqrt{\dfrac{s \left(s - c\right)^2}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}}\right] \\\\
& = 2 \left(s - a\right) \times \sqrt{\dfrac{s}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}} \times \left(s - b + s - c\right) \\\\
& = 2 \left(s - a\right) \times \left(2s - b - c\right) \times \sqrt{\dfrac{s}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}} \\\\
& = 2 \times \left(a + b + c - b - c\right) \times \sqrt{\dfrac{s \left(s - a\right)^2}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}} \;\; [\text{by equation (2a)}] \\\\
& = 2 a \sqrt{\dfrac{s \left(s - a\right)}{\left(s - b\right) \left(s - c\right)}} \\\\
& = 2 a \cot \left(\dfrac{A}{2}\right) = RHS
\end{aligned}$
Hence proved.