$\text { If } \sin y=x(\cos (a+y)) \text {, then find } \dfrac{d y}{d x} \text { when } x=0$

Given the equation $\sin y = x \cos(a + y)$, we can express $x$ as a function of $y$:

$x = \dfrac{\sin y}{\cos(a + y)}$

Differentiating both sides with respect to $y$ using the quotient rule:

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

Using the trigonometric identity $\cos A \cos B + \sin A \sin B = \cos(A - B)$, we get:

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

Therefore, $\dfrac{dy}{dx} = \dfrac{\cos^2(a + y)}{\cos a}$.

When $x = 0$, substituting into the original equation $\sin y = x \cos(a + y)$ gives $\sin y = 0$, which implies $y = 0$.

Substituting $y = 0$ into the expression for $\dfrac{dy}{dx}$:

$\dfrac{dy}{dx} = \dfrac{\cos^2(a + 0)}{\cos a} = \dfrac{\cos^2 a}{\cos a} = \cos a$

Answer: cos a