If xcos(a+y)=cosy, then prove that dydx=cos²(a+y)sina. Show that sinad²ydx²+sin²(a+y)dydx=0.

xcos(a+y)=cosy,
Which implies x=cosy/cos(a+y),
Now differentiate on both sides , we get
dx=[cos(a+y)(-siny)-cosy(-sin(a+y))]cos²(a+y)dy=sin(a+y-y)cos²(a+y)dy=sinacos²(a+y)dy
⇒dydx=cos²(a+y)sina
⇒d²ydx²=2cos(a+y)(-sin(a+y)sina×dydx)=-sin2(a+y)sina×dydx
Therefore,sinad²ydx²+sin²(a+y)dydx=0