If xsin(a+y) + sin(a)cos(a+y) = 0, prove that dy/dx = sin²(a+y)/sin(a)

Givenxsin(a+y)+sinacos(a+y)=0⇒x=−sinacos(a+y)sin(a+y)i.e.,dxdy=−sina×ddy(cos(a+y)sin(a+y))...[Differentiating w.r.t y both the sides]dxdy=−sina×sin(a+y)ddy(cos(a+y))−cos(a+y)ddy(sin(a+y))[sin(a+y)]2dxdy=−sina×−sin(a+y)sin(a+y)−cos(a+y)cos(a+y)[sin(a+y)]2=sina×[sin2(a+y)+cos2(a+y)[sin(a+y)]2]dxdy=sina×1sin2(a+y)Therefore,dydx=sin2(a+y)sinaHence proved.