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#
