In any ΔABC, prove that

Left hand side,

$a c \cos B-b c \cos A$

$=a c \frac{a^{2}+c^{2}-b^{2}}{2 a c}-b c \frac{b^{2}+c^{2}-a^{2}}{2 b c}\left[A s, \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \& \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right]$

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

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

$=\frac{2\left(a^{2}-b^{2}\right)}{2}$

$=a^{2}-b^{2}$

$=$ Right hand side. [Proved]