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

Steps to prove  sin(A+B)=2ab/(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

=sin(A+B)

=sin2(A+B)/2

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

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

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

=R.H.S

Hence  sin(A+B)=2ab/(b^2+a^2) is proved.

Detail Information:-

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