If sin x sqrt{cos y} - cos y sqrt{sin x} = 0, | vedclass

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Question

(C) Given the equation $\sin x \sqrt{\cos y} - \cos y \sqrt{\sin x} = 0$. Rearranging the terms,we get $\sin x \sqrt{\cos y} = \cos y \sqrt{\sin x}$.

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) Given the equation $\sin x \sqrt{\cos y} - \cos y \sqrt{\sin x} = 0$. Rearranging the terms,we get $\sin x \sqrt{\cos y} = \cos y \sqrt{\sin x}$. Squaring both sides,we obtain $\sin^2 x \cos y = \cos^2 y \sin x$. Assuming $\sin x 
eq 0$ and $\cos y 
eq 0$,we can divide by $\sin x \cos y$ to get $\frac{\sin x}{\cos y} = \frac{\cos y}{\sin x}$,which implies $\sin^2 x = \cos^2 y$. Thus,$\sin x = \cos y$ or $\sin x = -\cos y$. Taking $\sin x = \cos y$,we differentiate both sides with respect to $x$: $\frac{d}{dx}(\sin x) = \frac{d}{dx}(\cos y)$. This gives $\cos x = -\sin y \frac{dy}{dx}$. Therefore,$\frac{dy}{dx} = -\frac{\cos x}{\sin y}$. Since $\cos y = \sin x$,we have $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - \sin^2 x} = \cos x$. Substituting this back,$\frac{dy}{dx} = -\frac{\cos x}{\cos x} = -1$.

If $\sin x \sqrt{\cos y} - \cos y \sqrt{\sin x} = 0$,then $\frac{dy}{dx} = $

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