Solve this

Here,

$\cos y=x \cos (a+y)$, where $\cos a \neq \pm 1$

Differentiating both sides with respect to $x$, we get

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

$\frac{d y}{d x}[x \sin (a+y)-\sin y]=\cos (a+y)$

$\frac{d y}{d x}=\frac{\cos (a+y)}{x \sin (a+y)-\sin y}$

Multiplying the numerator and the denominator by $\cos (a+y)$ on th RHS we have,

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

$\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\cos y \sin (a+y)-\cos (a+y) \sin y}[$ Given $\cos y=x \cos (a+y)]$

$\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin [(a+y)-y]}[\because \sin (a-b)=\sin a \cos b-\cos a \sin b]$

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

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

Hence Proved.