(m2−n2)
=(sinAsinB)2−(cosAcosB)2
=(sin2Asin2B)−(cos2Acos2B)
⇒(m2−n2)sin2B=(sin2A−cos2Atan2B)
=1−cos2A−cos2Atan2B
=1−cos2A(1+tan2B)
=1−cos2Asec2B
=1−cos2Acos2B
=1−(cosAcosB)2
=1−n2
if cosA+sinB=m and sinA+cosB=n, prove that 2sin(A+B)=m2+n2-2