# Question #0e494

Nov 23, 2016

For $\Delta A B C$

we have $A + B + C = \pi$

So $\cot \left(A + B\right) = \cot \left(\pi - C\right)$

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

$\implies \cot A \cot B + \cot B \cot C + \cot C \cot A = 1 \text{ } \textcolor{red}{\left[1\right]}$

Now it is given

$\cot A + \cot B + \cot C = \sqrt{3}$

Squaring both sides we get

${\cot}^{2} A + {\cot}^{2} B + {\cot}^{2} C + 2 \left(\cot A \cot B + \cot B \cot C + \cot C \cot A\right) = 3 \text{ } \textcolor{red}{\left[2\right]}$

Now subtracting thrice of  from  we get

${\cot}^{2} A + {\cot}^{2} B + {\cot}^{2} C - \left(\cot A \cot B + \cot B \cot C + \cot C \cot A\right) = 0$

$\implies 2 {\cot}^{2} A + 2 {\cot}^{2} B + 2 {\cot}^{2} C - 2 \cot A \cot B - 2 \cot B \cot C - 2 \cot C \cot A = 0$

$\implies {\left(\cot A - \cot B\right)}^{2} + {\left(\cot B - \cot C\right)}^{2} + {\left(\cot C - \cot A\right)}^{2} = 0$

Now sum of three squared terms each of which is positive being zero each will be equal to zero.

Hence

$\cot A = \cot B = \cot C$

$\implies A = B = C$

This implies that $\Delta A B C$ is equilateral.