If CosA + SinB = m And SinA + CosB = n, prove that 2 Sin(A +B) = m2 + n2 – 2.
Given cosA + sinB = m And sinA + cosB = n
RHS = m2 + n2 – 2
=(cosA + sinB)2 +(sinA + cosB)2 – 2
= cos2A + sin2B + 2 cosA sinB + sin2A + cos2B + 2 sinA cosB – 2
= 1 + 1 + 2(cosA sinB + sinA cosB) - 2
We know that sin(A -B) = sinA cosB - cosA sinB
= 2 sin(A +B)
= LHS
Hence, proved.