The discussion revolves around a trigonometric identity related to the properties of triangles, specifically involving the cotangent of half-angles and the sides of triangle ABC.
The discussion is ongoing, with participants sharing their attempts and questioning the effectiveness of their methods. Some guidance has been offered regarding the use of the sine rule and the relationship between the angles of the triangle.
Participants note the importance of adhering to the problem's constraints and express concern about maintaining the integrity of the discussion format.
Wrichik Basu said:In any triangle ABC, prove that $$(b+c-a) \left( \cot {\frac {B}{2}} + \cot {\frac {C}{2}} \right)=2a \cot {\frac {A}{2}} $$
Buffu said:Did you try to solve this ?
Also you should not vandalise the template.
Use sine rule and A + B + C = 180.Wrichik Basu said:I tried in several ways, trying to change a, b, c to 2R sin A and like that, or trying the formulae for cot A/2, but got nowhere.