If $\cos y = x \cos (a+y)$ with $\cos a \neq \pm 1,$ prove that $\frac{dy}{dx} = \frac{\cos^2 (a+y)}{\sin a}$.

Given that,$\cos y = x \cos (a+y)$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\cos y) = \frac{d}{dx}(x \cos (a+y))$
Using the chain rule and product rule:
$-\sin y \frac{dy}{dx} = \cos (a+y) \cdot (1) + x \cdot (-\sin (a+y)) \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$x \sin (a+y) \frac{dy}{dx} - \sin y \frac{dy}{dx} = \cos (a+y)$
$\frac{dy}{dx} [x \sin (a+y) - \sin y] = \cos (a+y) \quad \dots(1)$
From the original equation,$x = \frac{\cos y}{\cos (a+y)}$. Substituting this into equation $(1)$:
$\frac{dy}{dx} \left[ \frac{\cos y}{\cos (a+y)} \sin (a+y) - \sin y \right] = \cos (a+y)$
$\frac{dy}{dx} \left[ \frac{\cos y \sin (a+y) - \sin y \cos (a+y)}{\cos (a+y)} \right] = \cos (a+y)$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$\frac{dy}{dx} \left[ \frac{\sin(a+y-y)}{\cos (a+y)} \right] = \cos (a+y)$
$\frac{dy}{dx} \left[ \frac{\sin a}{\cos (a+y)} \right] = \cos (a+y)$
$\frac{dy}{dx} = \frac{\cos^2 (a+y)}{\sin a}$.
Hence,proved.