In any $\triangle ABC$, prove that the area of the triangle $\Delta = \dfrac{b^2 + c^2 - a^2}{4 \cot A}$
$\begin{aligned}
\text{Area of } \triangle ABC & = \dfrac{1}{2} b c \sin A \\\\
& = \dfrac{1}{2} b c \times \left(\dfrac{\sin A}{\cos A}\right) \times \cos A \\\\
& = \dfrac{1}{2} b c \tan A \cos A \\\\
& = \dfrac{b c \cos A}{2 \cot A} \\\\
& \left[\text{Cosine rule: } \cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}\right] \\\\
& = \dfrac{b c \times \left(\dfrac{b^2 + c^2 - a^2}{2 bc}\right)}{2 \cot A} \\\\
& = \dfrac{b^2 + c^2 - a^2}{4 \cot A}
\end{aligned}$
Hence proved.