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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation $\sin y = x \cos (a + y).$ We can express $x$ as $x = \frac{\sin y}{\cos (a + y)}.$ Differentiating both sides with respect to $y$ using the quotient rule $\frac{d}{dy} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dy} - u \frac{dv}{dy}}{v^2}$: $\frac{dx}{dy} = \frac{\cos (a + y) \cdot \cos y - \sin y \cdot (-\sin (a + y))}{\cos^2 (a + y)}$ $\frac{dx}{dy} = \frac{\cos (a + y) \cos y + \sin y \sin (a + y)}{\cos^2 (a + y)}$ Using the trigonometric identity $\cos A \cos B + \sin A \sin B = \cos (A - B)$: $\frac{dx}{dy} = \frac{\cos (a + y - y)}{\cos^2 (a + y)} = \frac{\cos a}{\cos^2 (a + y)}$ Therefore,$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{\cos^2 (a + y)}{\cos a}.$ Thus,the correct option is $A$.

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

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