Question:
In ∆ABC, prove the following:
$b(c \cos A-a \cos C)=c^{2}-a^{2}$
Solution:
Let ABC be any triangle.
Consider
$b(c \cos A-a \cos C)=b c \cos A-a b \cos C$
$=b c\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)-a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)$
$=\frac{b^{2}+c^{2}-a^{2}-a^{2}-b^{2}+c^{2}}{2}$
$=\frac{2\left(c^{2}-a^{2}\right)}{2}$
$=c^{2}-a^{2}$
Hence proved.