Answer :

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.


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