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Question

If cos A +sinB =m and sinA+cosB=n prove that 2sin(A+B)=m²+n²-2

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Solution

Lets take m and = n and substitute on RHS

m2 + n2 - 2 = ( cosA + sinB)2+ ( sinA + cosB)2 =

cos2A + 2cosAsinB + sin2B +sin2A +2sinAcosB + cos2B -2

=(sin2A+cos2A )+ (sin2B+cos2B)-2 +2(cosAsinB +sinAcosB)
( Here Note that sinAcosB+cosAsinB = sin(A+B) and sin2x+cos2x = 1 )
= 1+1-2 + 2(cosAsinB +sinAcosB ) =2sin(A+B) =LHS

So LHS = RHS ; Hence Proved

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