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

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

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

Given: $\sin y = x \cos(a + y)$

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

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

From $\sin y = x \cos(a + y)$, we get $x = \frac{\sin y}{\cos(a + y)}$

$\frac{dy}{dx}\left[\cos y + \frac{\sin y}{\cos(a + y)}\sin(a + y)\right] = \cos(a + y)$

Using the identity:

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

$\frac{dy}{dx} \frac{\cos a}{\cos(a + y)} = \cos(a + y)$

Final Answer:

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