Given:
x cos(a+y)=cos y
∴x=cos ycos(a+y) .............(1)
Differentiating equation (1) w.r.t.y
⇒dxdy=−sin y[cos(a+y)]−cos y[−sin(a+y)]cos2(a+y)
⇒dxdy=−sin y(cos(a+y))+sin(a+y).cos ycos2(a+y)
⇒dxdy=sin(a+y).cos y−sin y.cos(a+y)cos2(a+y)
⇒dxdy=sin(a+y−y)cos2(a+y)
⇒dxdy=sin acos2(a+y)
⇒dydx=cos2(a+y)sin a .................(2)
Now we have sin adydx=cos2(a+y) ...............(3)
Differentiating equation (3) w.r.t. x, we will get
sin ad2ydx2=2cos(a+y)[−sin(a+y)]dydx
⇒sin ad2ydx2=−2 sin(a+y)cos(a+y).dydx
⇒sin ad2ydx2=−sin 2(a+y)dydx
⇒sin ad2ydx2+sin 2(a+y)dydx=0
Hence Proved.