Question 1017353
We use the identity that if A, B, C are angles of a triangle, then *[tex \large \tan A + \tan B + \tan C = \tan A \tan B \tan C]


Dividing both sides by *[tex \large \tan A \tan B \tan C], we obtain


*[tex \large \cot A \cot B + \cot B \cot C + \cot C \cot A = 1] (1)


The left-hand side equals *[tex \large \frac{1}{2}\left( (\cot A + \cot B + \cot C)^2 - \cot^2 A - \cot^2 B - \cot^2 C \right)], so after some rearranging we get:


*[tex \large (\cot A + \cot B + \cot C)^2 = 2 + \cot^2 A + \cot^2 B + \cot^2 C]


Using the fact that *[tex \large \cot A + \cot B + \cot C = \sqrt{3}]:


*[tex \large 3 = 2 + \cot^2 A + \cot^2 B + \cot^2 C \Rightarrow \cot^2 A + \cot^2 B + \cot^2 C = 1] (2)


(2)*2 + (1)*-2 yields the nice expression:


*[tex \large (\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2 = 0]


which implies that *[tex \large \cot A = \cot B = \cot C] or *[tex \large \tan A = \tan B = \tan C]. Therefore, *[tex \large A = B = C = 60^{\circ}].