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(A) Given the equation $x \cos (k+y)=\cos y$. We can write $x = \frac{\cos y}{\cos (k+y)}$. Differentiating both sides with respect to $y$: $\frac{dx}

Vedclass · Accepted Answer

$\sin k$

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(A) Given the equation $x \cos (k+y)=\cos y$. We can write $x = \frac{\cos y}{\cos (k+y)}$. Differentiating both sides with respect to $y$: $\frac{dx}{dy} = \frac{\cos(k+y) \cdot (-\sin y) - \cos y \cdot (-\sin(k+y))}{\cos^2(k+y)}$ $\frac{dx}{dy} = \frac{\sin(k+y)\cos y - \cos(k+y)\sin y}{\cos^2(k+y)}$ Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, we get: $\frac{dx}{dy} = \frac{\sin(k+y-y)}{\cos^2(k+y)} = \frac{\sin k}{\cos^2(k+y)}$. Therefore, $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{\cos^2(k+y)}{\sin k}$. Now, substitute $y = \frac{\pi}{2}$: $\frac{dy}{dx} = \frac{\cos^2(k+\frac{\pi}{2})}{\sin k} = \frac{(-\sin k)^2}{\sin k} = \frac{\sin^2 k}{\sin k} = \sin k$. Thus, option $A$ is correct.

If $x \cos (k+y)=\cos y$, then $\frac{dy}{dx}$ at $y=\frac{\pi}{2}$ is

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