Question

Prove that :-
$\frac{\cos A}{\left(c\cos B+b\cos C\right)}+\frac{\cos B}{\left(a\cos C+\cos A\right)}+\frac{\cos C}{\left(a\cos B+b\cos A\right)}=\frac{a^2+b^2+c}{2abc}$

Answer 1

Using formulae $\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos B=\frac{a^2+c^2-b^2}{ac}$

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

putting these formulae in $L.H.S$

We have

$=\frac{\frac{b^2+c^2-a^2}{2bc}}{a\left(\frac{a^2+c^2-b^2}{2ac}\right)+b\left(\frac{a^2+b^2-c^2}{2ab}\right)}+\frac{\frac{a^2+c^2-b^2}{2ac}}{a\left(\frac{a^2+b^2-c^2}{2ab}\right)+c\left(\frac{b^2+c^2-a^2}{2bc}\right)}+\frac{\frac{a^2+b^2-c^2}{2ab}}{a\left(\frac{a^2+c^2-b^2}{2ac}\right)+b\left(\frac{b^2+c^2-a^2}{2bc}\right)}$

$=\frac{\frac{b^2+c^2-a^2}{bc}}{\frac{2a^2}{a}}+\frac{\frac{a^2+c^2-b^2}{ac}}{\frac{2b^2}{b}}+\frac{\frac{a^2+b^2-c^2}{ab}}{\frac{2c^2}{c}}$

$\Rightarrow \frac{a^2+b^2+c^2}{2abc}=R.H.S$

Hence proved