If siny = xsin(a+y), prove that dydx = sin2(a+y)/sina

siny=xsin(a+y) ⇒x=sinysin(a+y)
Differentiating with sides w.r.t.y, we get
dxd y=cosysin(a+y)−sinycos(a+y)sin2(a+y)[Quotient Rule]
⇒dxdy=cosy(sinacosy+ cosasiny)−siny(cosacosy−sinasiny)sin2(a+y)
⇒dxdy=cos2ysina+ cosasinycosy−sinycosacosy+sinasin2ysin2(a+y)
⇒dxdy=cos2ysina+sinasin2ysin2(a+y)
⇒dxdy=sina(cos2y+sin2y)sin2(a+y)
⇒dxdy=sinasin2(a+y)
⇒dydx=sin2(a+y)sina
Hence Proved.