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

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

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

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

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