Let  $ABC$ be a triangle such that 

$\cot A+\cot B+\cot C=\sqrt{3}$ then prove that $ABC$  is an equilateral 

$A+B+C=\pi ; A+B=\pi -C$

Apply 'Cot' on both sides 

$\mathrm{Cot}(A+B)=\mathrm{Cot}(\pi -C)$

$\frac{\mathrm{Cot}A \mathrm{Cot}B-1}{\mathrm{Cot}B+\mathrm{Cot}A}=-\mathrm{Cot}C$

$\Rightarrow \mathrm{Cot}A \mathrm{Cot}B-1=-\mathrm{Cot}C \mathrm{Cot} B-\mathrm{Cot} C \mathrm{Cot} A$

$\Rightarrow \mathrm{Cot}A \mathrm{Cot}B+\mathrm{Cot}B \mathrm{Cot}C+\mathrm{Cot}C \mathrm{Cot}A=1$

$\sum \mathrm{Cot} A \mathrm{Cot} B=1\to \left(1\right)$

Given $\mathrm{Cot}A+\mathrm{Cot}B+\mathrm{Cot}C=\sqrt{3}\to \left(2\right)$

Let $\mathrm{Cot}A=x$  $\mathrm{Cot}B=y$   $\mathrm{Cot}C=z$

Then $x+y+z=\sqrt{3}$ and $xy+yz+zx=1\to \left(3\right)$

Now $(x-y{)}^2+(y-z{)}^2+(z-x{)}^2$

$=2\left(x^2+y^2+z^2\right)-2(xy+yz+zx)$

$=2\left[(x+y+z{)}^2-2(xy+yz+zx)\right]-2(xy+yz+zx)$

$=2\left[(\sqrt{3}{)}^2-2(1)\right]-2$  from (3)

$=2(3-2)-2=2-2=0$

$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=0$

$\therefore$ $(x-y{)}^2=0,(y-z{)}^2=0$ and $(z-x{)}^2=0$

$\Rightarrow x-y=0, y-z=0, z-x=0\Rightarrow x=y=z$

But $x+y+z=\sqrt{3}$  $\Rightarrow x=y=z=\frac{1}{\sqrt{3}}$

$\Rightarrow \mathrm{Cot}A=\mathrm{Cot}B=\mathrm{Cot}C=\frac{1}{\sqrt{3}}$

Since $A+B+C={180}^{\circ }$ $\Rightarrow A=B=C={60}^{\circ }$

Since all the three angles are equal, it is an equilateral triangle.