If sinA+sinB=a and cosA+cosB=b than prove that cos(A+B)=(b^2-a^2)/(b^2+a^2)

Steps to prove cos(A+B)=(b^2-a^2)/(b^2+a^2)

sinA+sinB=a-(I)

cosA+cosB=b-(II)

Dividing equation (I) and (II)

⇒(sinA+sinB)/(cosA+cosB)=a/b

⇒{2sin(A+B)/2.cos(A-B)/2}/{2cos(A+B)/2.cos(A-B)/2}=a/b

⇒{sin(A+B)/2}/cos(A+B)/2}=a/b

⇒tan(A+B)/2=a/b

Take L.H.S

=cos(A+B)

=cos2(A+B)/2

={1-tan^2(A+B)/2}/{1+tan^2(A+B)/2}

={1-(a^2/b^2)}/{1+(a^2/b^2)}

={(b^2-a^2)/b^2}/{(b^2+a^2)/b^2}

=(b^2-a^2)/(b^2+a^2)

=R.H.S  {Proved}

Detail Information:-

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