In any ΔABC, prove that
Question:

In any ΔABC, prove that

$a c \cos B-b c \cos A=\left(a^{2}-b^{2}\right)$

 

Solution:

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]

 

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