Question

Prove that $(b-c)\cot \frac{A}{2}+(c-a)\cot \frac{B}{2}+(a-b)]cot\frac{C}{2}=0$.

Answer 1

In a triangle,

$\tan \frac{A}{2}=\frac{\Delta }{s(s-a)}$

$\tan \frac{B}{2}=\frac{\Delta }{s(s-b)}$

$\tan \frac{C}{2}=\frac{\Delta }{s(s-c)}$

We have,

$\Rightarrow \frac{s}{2\Delta }\left(\right(b-c\left)\right(b+c-a)+(c+a-b\left)\right(c-a)+(a-b\left)\right(a+b-c\left)\right)=a^2+b^2+c^2-a^2-b^2-c^2-a(b-c)-b(c-a)-c(a-b)=0$

LHS=RHS