In any ΔABC, prove that
$\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=\frac{\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)}{2 \mathrm{abc}}$
Need to prove: $\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=\frac{\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)}{2 \mathrm{abc}}$
Left hand side
$=\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}$
$=\frac{\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{a}^{2}}{2 \mathrm{abc}}+\frac{\mathrm{c}^{2}+\mathrm{a}^{2}-\mathrm{b}^{2}}{2 \mathrm{abc}}+\frac{\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}}{2 \mathrm{abc}}$
$=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}{2 \mathrm{abc}}$
$=$ Right hand side. [Proved]