Question

Prove (b^2-c^2)/a*cosA + (c^2-a^2)/b*cosB + (a^2-b^2)/c*cosC =0

Answer 1

In a triangle $\Delta ABC$, the cosine law is

$\cos A=\frac{b^2+c^2-a^2}{2bc} \\ \cos B=\frac{c^2+a^2-b^2}{2ca} \\ \cos C=\frac{a^2+b^2-c^2}{2ab}$

Now,

$\frac{(b^2-c^2)}{a} \cos A=\frac{(b^2-c^2)}{a} \frac{b^2+c^2-a^2}{2bc}\\ \frac{(b^2-c^2)}{a} \cos A=\frac{b^4-c^4-a^2(b^2-c^2)}{2abc}$

Similarly,