If sin y = x sin (a + y), then frac{dy}{dx} = | vedclass

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(C) Given the equation: $\sin y = x \sin (a + y)$We can express $x$ as: $x = \frac{\sin y}{\sin (a + y)}$Differentiating both sides with respect to $x

Vedclass · Accepted Answer

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

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(C) Given the equation: $\sin y = x \sin (a + y)$We can express $x$ as: $x = \frac{\sin y}{\sin (a + y)}$Differentiating both sides with respect to $x$ using the quotient rule:$\frac{dx}{dx} = \frac{d}{dx} \left[ \frac{\sin y}{\sin (a + y)} \right]$$1 = \frac{\cos y \cdot \frac{dy}{dx} \cdot \sin (a + y) - \sin y \cdot \cos (a + y) \cdot \frac{dy}{dx}}{\sin^2 (a + y)}$Factor out $\frac{dy}{dx}$ in the numerator:$1 = \frac{\frac{dy}{dx} [\sin (a + y) \cos y - \cos (a + y) \sin y]}{\sin^2 (a + y)}$Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$1 = \frac{\frac{dy}{dx} \cdot \sin (a + y - y)}{\sin^2 (a + y)}$$1 = \frac{\frac{dy}{dx} \cdot \sin a}{\sin^2 (a + y)}$Therefore,$\frac{dy}{dx} = \frac{\sin^2 (a + y)}{\sin a}$.

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

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