Question
In ∆ABC prove the following : $a c \cos B-b c \cos A=\left(a^2-b^2\right)$.
✓
Answer
LHS = ac cos B – bc cos A
$\begin{aligned} & =a c\left(\frac{c^2+a^2-b^2}{2 c a}\right)-b c\left(\frac{b^2+c^2-a^2}{2 b c}\right) \\ & =\frac{1}{2}\left(c^2+a^2-b^2\right)-\frac{1}{2}\left(b^2+c^2-a^2\right)\end{aligned}$
$\begin{aligned} & =\frac{1}{2}\left(c^2+a^2-b^2-b^2-c^2+a^2\right) \\ & =\frac{1}{2}\left(2 a^2-2 b^2\right)=a^2-b 2=\text { RHS. }\end{aligned}$
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