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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Similar questions

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

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

m = \frac{sin A}{sin B}, n = \frac{cos A}{cos B}

(m^{2} - n^{2}) {sin}^{2} B = 1 - n^{2}

c o s B / s i n A = n

c o s B / c o s A = m

(m^{2} + n^{2}) c o s^{2} A = n^{2}

\frac{sin A - sin B}{cos A + cos B} + \frac{cos A - cos B}{sin A + sin B} = 0

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