Question

prove that in a triangle
$a(b\cos c-c\cos B)=b^2-c^2$

Answer 1

$c^2=a^2+c^2-2abcosB$

converting equation in law of cosines

$b^2=a^2+c^2-2accosB$

$c^2=a^2+b^2-2abcosC$

Second equation is subtracted from the first one and the result came

$b^2-c^2=c^2-b^2-2accosB+2abcosC$

$c^2$and $-b^2$ goes on the other side of the = gives

$2b^2-2c^2=-2accosB+2abcosC$

$c^2-b^2$ come in brackets and 2 is taken as common gives

$2(b^2-c^2)=2a(-ccosB+bcosC)$

So $b^2-c^2=a(bcosC-ccosB)$