In $\triangle ABC$, with usual notations, if $\Delta$ denotes the area of triangle $ABC$ then the value of $2s(b+c-a)\tan\left(\dfrac{A}{2}\right)$ is equal to $\ldots$

We know that the semi-perimeter of a triangle is given by $s = \dfrac{a+b+c}{2}$, which implies $2s = a+b+c$.

The term $b+c-a$ can be rewritten as:

$b+c-a = (a+b+c) - 2a = 2s - 2a = 2(s-a)$

Substituting this into the given expression, we get:

$2s(b+c-a)\tan\left(\dfrac{A}{2}\right) = 2s(2(s-a))\tan\left(\dfrac{A}{2}\right) = 4s(s-a)\tan\left(\dfrac{A}{2}\right)$

Using the half-angle formula for tangents in a triangle:

$\tan\left(\dfrac{A}{2}\right) = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$

Multiplying the numerator and the denominator inside the square root by $s(s-a)$ gives:

$\tan\left(\dfrac{A}{2}\right) = \dfrac{\sqrt{s(s-a)(s-b)(s-c)}}{s(s-a)}$

By Heron's formula, the area of the triangle is $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$. Therefore:

$\tan\left(\dfrac{A}{2}\right) = \dfrac{\Delta}{s(s-a)}$

Substituting this back into the modified expression:

$4s(s-a)\tan\left(\dfrac{A}{2}\right) = 4s(s-a) \times \dfrac{\Delta}{s(s-a)} = 4\Delta$

Answer: $4\Delta$