Question

If A, B, C are the angles of a triangle, show that if in $\Delta$ ABC, $\cot A+\cot B+\cot C=\sqrt{3}$, prove that the triangle is equilateral.

Answer 1

Squaring the given relation, we have
$\sum {\cot }^{2}A+2\sum \cot A\cot B=3\sum \cot A\cot B$
because $\sum {\cot }^{2}A-\sum \cot A\cot B=0$
Above is of form $x^2+y^2+z^2-xy-yz-zx=0$
or $\frac{1}{2}\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]=0$
Above is possible only when $x-y=0,y-z=0$, $z-x=0$
or $x=y=z$ or $\cot A=\cot B=\cot C$
$\therefore A=B=C\Rightarrow \Delta$ is equilateral.