Given, in $\triangle A B C$,
$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\,\,\,\,\,\,\,\,...(i)$
From sine rule,
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\,\,\,\,\,\,\,\,...(ii)$
On dividing Eq. (ii) by Eq. (i), we get
$\Rightarrow \tan A=\tan B=\tan C $
$\Rightarrow A=B=C$
So, $\triangle A B C$ is an equilateral triangle
$\begin{pmatrix}\because \text { in } \Delta A B C, & A+B+C=180^{\circ} \\ & \Rightarrow A+A+A=180^{\circ} \\ & \Rightarrow \quad A=60^{\circ}=B=C\end{pmatrix}$
Area of equilateral $\triangle A B C$
$=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4}(2)^{2} \,\,\,\,(\because a=2, $ given $) $
$=\frac{\sqrt{3}}{4} \times 4=\sqrt{3}$