Trig2.10

In a triangle ABC if $\cot A+\cot B+\cot C=\sqrt{3}\,$ then show that the triangle is equilateral.

Given $\cot A+\cot B+\cot C=\sqrt{3}\,$ Let $\cot A=a, \cot B=b, \cot C=c\,$

Then $a+b+c=\sqrt{3}\,$ Let this be equation 1.

Since $A+B+C=180^\circ,A+B=180^\circ-C\,$

Therefore, $\cot (A+B)=\cot (180-C)\,$

$\frac{\cot A \cot B-1}{\cot B \cot B+\cot A}=-\cot C\,$

$\cot A \cot B+\cot B \cot C+\cot C \cot A=1\,$ implies $ab+bc+ca=1\,$

From 1 and 2

$(a+b+c)^2-3(ab+bc+ca)=0\,$

$a^2+b^2+c^2-ab-bc-ca=0\,$

$(a-b)^2+(b-c)^2+(c-a)^2=0\,$ implies $a=b=c\,$

Hence $\cot A=\cot B=\cot C\,$ implies $A=B=C\,$

Therefore triangle ABC is equilateral.