If y sqrt{1-x^{2}}+x sqrt{1-y^{2}}=1,then fra | vedclass

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(A) Given the equation $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$. Substitute $x = \sin \alpha$ and $y = \sin \beta$,where $\alpha = \sin^{-1} x$ and $\bet

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$-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$

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(A) Given the equation $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$. Substitute $x = \sin \alpha$ and $y = \sin \beta$,where $\alpha = \sin^{-1} x$ and $\beta = \sin^{-1} y$. The equation becomes $\sin \beta \cos \alpha + \sin \alpha \cos \beta = 1$. Using the trigonometric identity $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$,we get $\sin(\alpha + \beta) = 1$. Thus,$\alpha + \beta = \sin^{-1}(1) = \frac{\pi}{2}$. Substituting back,we have $\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) = \frac{d}{dx}(\frac{\pi}{2})$. $\frac{1}{\sqrt{1-x^{2}}} + \frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = 0$. Rearranging the terms: $\frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^{2}}}$. Therefore,$\frac{dy}{dx} = -\sqrt{\frac{1-y^{2}}{1-x^{2}}}$.

If $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$,then $\frac{dy}{dx}=$

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