In a \(\Delta A B C\), if \(\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\) and \(a=2\), then its area is

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
\(\tan A=\tan B=\tan C\)
\(\Rightarrow \quad A=B=C\)
So, \(\triangle A B C\) is an equilateral triangle
\(\left(\begin{array}{rrr}
\because \text { in } \triangle A B C, & & A+B+C=180^{\circ} \\
& \Rightarrow & A+A+A=180^{\circ} \\
& \Rightarrow & A=60^{\circ}=B=C
\end{array}\right)\)
\(\therefore\) Area of equilateral \(\triangle A B C\)
\(\begin{aligned}
&=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4}(2)^{2} \quad(\because a=2, \text { given }) \\
&=\frac{\sqrt{3}}{4} \times 4=\sqrt{3}
\end{aligned}\)