Question

$a(b\cos C-c\cos B)=b^2-c^2$.

Answer 1

$a(b\cos C-c\cos B)$
$=a\{b.\frac{a^2+b^2-c^2}{2ab}-c.\frac{c^2+a^2-b^2}{2ca}\}$
$=\frac{1}{2}(a^2+b^2-c^2-c^2-a^2-c^2)=b^2-c^2$
Alt. On putting $a=b\cos C+c\cos B$,
$L.H.S=b^2{\cos }^{2}C-c^2{\cos }^{2}B$
$=b^2(1-{\sin }^{2}C)-c^2(1-{\sin }^{2}B)=b^2-c^2$
$\because b\sin C=c\sin B$
By sine formula.