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