Question

in any triangle abc, prove that a(bcosC -ccosB)= b2-c2

Answer 1

Answer:

LHS. =a(bcosC−(c)cosB)

=abcosC−a(c)cosB

=ab.a2+b2−c22ab−ac.c2+a2−b22ca

=a2+b2−c22−c2+a2−b22

=a2+b2−c2−c2+a2−b22

=2b2−c22

=b2−c2= RHS Hence Proved.