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

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

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

$\cos y=x \cos (a+y)$

so $x=\frac{\cos y}{\cos (a+y)}⇒\frac{dx}{dy}=\frac{\sin y\cos(a+y)-\cos y\sin(a+y)}{\cos^2 (a+y)}$

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