If $x\, cos (a+y)+cos\, a\, sin (a+y) =0, $ then $\frac{dy}{dx}$ is equal to

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

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

The correct answer is Option (2) → $-\frac{cos^2(a+y)}{cos\, a}$ $x\cos (a+y)=-\cos a\sin (a+y)$ $x=-\cos a\frac{\sin(a+y)}{\cos(a+y)}$ differentiating wrt x $\frac{dx}{dy}=-\cos a\frac{(\cos^2(a+y)+\sin^2(a+y))}{\cos^2(a+y)}$ $⇒\frac{dy}{dx}=-\frac{\cos^2(a+y)}{\cos a}$