If sqrt{x-xy} + sqrt{y-xy} = 1,then frac{dy}{ | vedclass

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(A) Given equation is $\sqrt{x(1-y)} + \sqrt{y(1-x)} = 1$. Let $x = \sin^2 	heta$ and $y = \sin^2 \phi$. Then $\sqrt{\sin^2 	heta (1-\sin^2 \phi)} +

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$-\sqrt{\frac{y-y^2}{x-x^2}}$

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(A) Given equation is $\sqrt{x(1-y)} + \sqrt{y(1-x)} = 1$. Let $x = \sin^2 	heta$ and $y = \sin^2 \phi$. Then $\sqrt{\sin^2 	heta (1-\sin^2 \phi)} + \sqrt{\sin^2 \phi (1-\sin^2 	heta)} = 1$. $\sin 	heta \cos \phi + \sin \phi \cos 	heta = 1$. $\sin(	heta + \phi) = 1$. $	heta + \phi = \frac{\pi}{2}$. $\phi = \frac{\pi}{2} - 	heta$. Since $y = \sin^2 \phi$,we have $y = \sin^2(\frac{\pi}{2} - 	heta) = \cos^2 	heta$. Thus $x = \sin^2 	heta$ and $y = \cos^2 	heta$. Adding these,$x+y = \sin^2 	heta + \cos^2 	heta = 1$,so $y = 1-x$. Differentiating with respect to $x$,$\frac{dy}{dx} = -1$. Alternatively,checking the options: $\frac{dy}{dx} = -\sqrt{\frac{y(1-y)}{x(1-x)}} = -\sqrt{\frac{(1-x)x}{x(1-x)}} = -\sqrt{1} = -1$. Thus,option $A$ is correct.

If $\sqrt{x-xy} + \sqrt{y-xy} = 1$,then $\frac{dy}{dx} = $

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