Question

Prove that

$\frac{cosA}{ccosB+bcosC}+\frac{cosB}{acosC+ccosA}+\frac{cosC}{acosB+bcosA}=\frac{a^2+b^2+c^2}{2abc}$

Answer 1

$LHS=\frac{\cos A}{c\cos B+b\cos C}+\frac{\cos B}{a\cos Cc\cos A}+\frac{\cos C}{a\cos Bb\cos A}$
From Cosine rule
$=\frac{\frac{b^2+c^2-a^2}{2bc}}{c\left(\frac{c^2+a^2-b^2}{2ac}\right)+b\left(\frac{a^2+b^2-c^2}{2ab}\right)}+\frac{\frac{a^2+c^2-b^2}{2ac}}{a\left(\frac{a^2+b^2-c^2}{2ab}\right)+c\left(\frac{b^2+c^2-a^2}{2bc}\right)}+\frac{\frac{a^2+b^2-c^2}{2ab}}{a\left(\frac{a^2+c^2-b^2}{2ac}\right)+b\left(\frac{b^2+c^2-a^2}{2bc}\right)}$
$=\frac{b^2+c^2-a^2}{2abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}$
$=\frac{a^2+b^2+c^2}{2abc}$

$=RHS$