If $\cos y=x \cos (\mathrm{a}+y)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

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

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

The correct answer is Option (1) - $\frac{\cos ^2(a+y)}{\sin a}$