In ∆ABC, prove the following:
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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