$2ab=2(\sin x+\sin y)(\cos x+\cos y)$
$\Rightarrow 2(\sin x\cos x+\sin x\cos y+\sin y\cos x+\sin y\cos y)$
$\Rightarrow \sin 2x+\sin 2y+2\sin(x+y)$
$\Rightarrow 2\sin(x+y)\cos(x-y)+2\sin (x+y)$
$\Rightarrow 2\sin(x+y)[\cos(x-y)+1]$------(1)
$a^2+b^2=(\sin x+\sin y)^2+(\cos x+\cos y)^2$
$\Rightarrow \sin^2x+\sin ^2y+2\sin x\sin y+\cos^2x+\cos ^2y+2\cos x\cos y$
$\Rightarrow (\sin ^2x+\cos^2x)+(\sin ^2y+\cos^2y)+2(\cos x\cos y+\sin x\sin y)$
$\Rightarrow 2+2\cos(x-y)$
$\Rightarrow 2(1+\cos(x-y)]$-------(2)
Dividing (1) by (2) we get
$\large\frac{2ab}{a^2+b^2}=\frac{2\sin(x+y)[\cos(x-y)+1]}{2[1+\cos(x-y)]}$
$\large\frac{2ab}{a^2+b^2}=$$\sin(x+y)$
Hence proved