if cosA+sinB=m and sinA+cosB=n, prove that 2sin(A+B)=m²+n²-2

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Solution

RHS= m^2+ n^2- 2 = ( cosA + sinB)^2+ ( sinA + cosB)^2 = cos^2A + 2cosAsinB + sin^2B +sin^2A +2sinAcosB + cos^2B -2 =(sin^2A+cos^2A )+ (sin^2B+cos^2B)-2 +2cosAsinB +2sinAcosB = 1+1-2 + 2(cosAsinB +sinAcosB ) =2sin(A+B) =LHS

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2 s i n (A + B) = m^{2} + n^{2} - 2

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