Given that,
rarrcosx+coy=a....[1]
rarrsinx+siny=b....[2]
Squaring and adding [1] and [2], we get,
rarrcos^2x+2cosxcosy+cos^2y+sin^2x+2sinxsiny+sin^2y=a^2+b^2
rarr2+2(cosxcosy+sinxsiny)=a^2+b^2
rarr2(1+cos(x-y))=a^2+b^2
rarrcos(x-y)=(a^2+b^2)/2-1
Dividing equation [1] by [2], we get,
rarr(cosx+cosy)/(sinx+siny)=a/b
rarr(2cos((x+y)/2)cos((x-y)/2))/(2sin((x+y)/2)cos((x-y)/2))=a/b
rarrcot((x+y)/2)=a/b
rarrtan((x+y)/2)=b/a
rarr(x+y)/2=tan^(-1)(b/a)
rarrx+y=2tan^(-1)(b/a)
As, 2tan^(-1)x=sin^(-1)((2x)/(1+x^2)),we have,
rarrx+y=sin^(-1)((2*(b/a))/(1+(b/a)^2))=sin^(-1)((2ab)/(a^2+b^2))
rarrsin(x+y)=(2ab)/(a^2+b^2)
Now,
LHS=sin2x+sin2y
=2sin(x+y)*cos(x-y)
=2[(2ab)/(a^2+b^2)][(a^2+b^2)/2-1]
=2ab[2/(a^2+b^2)*(a^2+b^2)/2-2/(a^2+b^2)]
=2ab[1-2/(a^2+b^2)]=RHS