91Ó°ÊÓ

Identity: $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^2+b^2+c^2}{2abc}$

Law of Cosines -

$a^2=b^2+c^2-2bc\cos A→1$

$b^2=a^2+c^2-2ac\cos B→2$

$c^2=a^2+b^2-2ab\cos C→3$

Now add$1,2and3$, we will get -

role="math" localid="1647111133522" $⇒a^2+b^2+c^2=b^2+c^2-2bc\cos A+a^2+c^2-2ac\cos B+a^2+b^2-2ab\cos C$

role="math" localid="1647111143924" $⇒a^2+b^2+c^2=2a^2+2b^2+2c^2-2bc\cos A-2ac\cos B-2ab\cos C$

role="math" localid="1647111153146" $⇒a^2+b^2+c^2+2bc\cos A+2ac\cos B+2ab\cos C=2a^2+2b^2+2c^2$

role="math" localid="1647111163322" $⇒2bc\cos A+2ac\cos B+2ab\cos C=2a^2+2b^2+2c^2-a^2-b^2-c^2$

role="math" localid="1647111170365" $⇒2\left(bc\cos A+ac\cos B+ab\cos C\right)=a^2+b^2+c^2$

role="math" localid="1647111180525" $⇒bc\cos A+ac\cos B+ab\cos C=\frac{a^2+b^2+c^2}{2}$

role="math" localid="1647111192253" $⇒\frac{bc\cos A+ac\cos B+ab\cos C}{abc}=\frac{a^2+b^2+c^2}{2abc}$

$⇒\frac{bc\cos A}{abc}+\frac{ac\cos B}{abc}+\frac{ab\cos C}{abc}=\frac{a^2+b^2+c^2}{2abc}$

Hence proved.