If $\cos y=x\cdot \cos (a+y)\text{then}\frac{dy}{dx}=$

$x=\frac{\cos y}{\cos (a+y)}$ and then differentiate

$$\begin{gathered} \text{ Differentiating w.r.t. }y\text{, we get } \\ \frac{dx}{dy}=\frac{d}{dy}\left(\frac{\cos y}{\cos (a+y)}\right) \end{gathered}$$

$$\begin{gathered} =\left(\frac{\cos (a+y)(-\sin y)-\cos y(-\sin (a+y\left)\right)}{{\cos }^2(a+y)}\right) \\ =\left(\frac{-\cos (a+y)\sin y+\cos y\sin (a+y)}{{\cos }^2(a+y)}\right) \end{gathered}$$

$=\left(\frac{\sin (a+y-y)}{{\cos }^2(a+y)}\right)$

$=\frac{\sin a}{{\cos }^2(a+y)}$

$\therefore \frac{dy}{dx}=\frac{{\cos }^2(a+y)}{\sin a}$