Question

answer this question of maths fast​

Answer 1

sin(A+B)sin(A-B)........

Answer 2

Given:

Sina sin(a+2b)-sinb sin(b+2a )

=2 {sina sin(a+2b)-sinbsin(b+2a)}/2

=cos(a+2b-a)-cos(a+2b+a)-{cos(b+2a-b)-cos(b+2a+b)}/2

Since,[2sinAcosB=cos(A-B)-cos(A+B)]

={cos(2b)-cos(2a+2b)-cos(2a)+cos(2a+2b)}/2

=(cos2b-cos2a)/2

Also,we know {cosA-cosB=-2 sin(A+B)/2*sin(a-b)/2

={-2 sin(2a+2b)/2*sin(2b-2a)/2}/2

={-2 sin(a+b)*(-)(a-b)}/2

=sin(a+b)*sin(a-b)

Hence, proved