prove that

I reached to a solution
a+b+c=180
a+b=180-c
tan(a+b)=tan(180-c) = -tan c
tan a + tan b + tan c = tan a tan b tan c
cot a cot b + cot a cot c + cot b cot c = 1
cot²a + cot²b + cot²c = 1
(cot a - 1/√3)²+(cot b -1/√3)²+(cot c -1/√3)²=0
hense abc is equalitrial triangle.

Is thair any different method for this.

In a triangle ABC, \sum_{cyc} \cot A \ge \sqrt{3} , where the equality occurs when A=B=C. Hence Proved that the above triangle is an equilateral triangle.

To Prove that \sum_{cyc} \cot A \ge \sqrt{3} :

You can use Jensen's Inequality I presume ie., 1/3 \sum_{cyc} \cot A \ge \cot(\frac{A+B+C}{3}) = \cot (\frac \pi3)= \frac {1}{\sqrt{3}} .