$\mathrm{In} ∆le{}_{ }\mathrm{ABC},\frac{\mathrm{a}}{\cos \mathrm{A}}=\frac{\mathrm{b}}{\cos \mathrm{B}}=\frac{\mathrm{c}}{\cos \mathrm{C}}; \mathrm{i}\text{⨍ b=2,}$ then the area of the triangle is

$$\begin{gathered} \frac{\mathrm{a}}{\cos \mathrm{A}}=\frac{\mathrm{b}}{\cos \mathrm{B}}=\frac{\mathrm{c}}{\cos \mathrm{C}}\Rightarrow \frac{\sin \mathrm{A}}{\cos \mathrm{A}}=\frac{\sin \mathrm{B}}{\cos B}=\frac{\sin \mathrm{C}}{\cos \mathrm{C}} \\ \Rightarrow \tan \mathrm{A}=\tan \mathrm{B}=\tan \mathrm{C} \\ \Rightarrow \mathrm{A}=\mathrm{B}=\mathrm{C}={60}^0\Rightarrow triangle is equilateral triangle \\ Given b=2\Rightarrow a=c=2 \end{gathered}$$

Area$=\frac{\sqrt{3}}{4}{{\left(a\right)}^2=\frac{\sqrt{3}}{4}\left(2\right)}^2=\sqrt{3}.$