Question

Prove that: $a\left(b\cos C-c\cos B\right)=b^2-c^2$

Answer 1

$L.H.S.$

$a(b\cos C-c\cos B)$

$=ab\cos C-ac\cos B$

Using cosine formula

$\cos C=\frac{a^2+b^2-c^2}{2ab}$ and $\cos B=\frac{a^2+c^2-b^2}{2ac}$

Then,

$=ab\left(\frac{a^2+b^2-c^2}{2ab}\right)-ac\left(\frac{a^2+c^2-b^2}{2ac}\right)$

$=\frac{a^2+b^2-c^2-a^2-c^2+b^2}{2}$

$=\frac{2(b^2-c^2)}{2}$

$=b^2-c^2$