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

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

Differentiating both sides with respect to $x$:

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

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

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

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

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Collecting all terms with $\frac{dy}{dx}$ on one side:

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

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

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

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From the original equation $\sin y = x \cos(a + y)$:

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

Substituting into the expression:

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

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

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

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Simplifying the complex fraction:

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

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

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Using the cosine addition formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$:

$\cos y \cos(a + y) + \sin y \sin(a + y) = \cos[(a + y) - y]$

$= \cos a$

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Therefore:

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