In ABC, prove that ac B − bc A = a^2 − b^2 - Vidyadip

Question

In $\triangle ABC$, prove that $ac\ \cos B − bc \cos\ A = a^2 − b^2$

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Answer

$ ac \cos B-b c \cos A$
$=a c\left(\frac{c^2+a^2-b^2}{2 a c}\right)-b c\left(\frac{b^2+c^2-a^2}{2 b c}\right) \ldots \ldots .[\text { By cosine rule }]$
$=\left(\frac{c^2+a^2-b^2}{2}\right)-\left(\frac{b^2+c^2-a^2}{2}\right)$
$=\frac{c^2+a^2-b^2-b^2-c^2+a^2}{2}$
$=\frac{2 a^2-2 b^2}{2}$
$=a^2-b^2$

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