Question

In a triangle $\Delta ABC$ prove that $a(b\cos C-c\cos B)=b^2-c^2$.

Answer 1

In any triangle $\Delta ABC$ we have, $\cos B=\frac{a^2+c^2-b^2}{2ab}$......(1) and $\cos C=\frac{a^2+c^2-b^2}{2ac}$........(2).

Now,

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

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

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

$=b^2-c^2$.