If sqrt{1-x^2}+sqrt{1-y^2}=a(x-y),then left[l | vedclass

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(B) Given,$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$.Let $x=\sin \alpha$ and $y=\sin \beta$. Then $\cos \alpha + \cos \beta = a(\sin \alpha - \sin \beta)$.Usi

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$x\left(1-y^2\right)$

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(B) Given,$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$.Let $x=\sin \alpha$ and $y=\sin \beta$. Then $\cos \alpha + \cos \beta = a(\sin \alpha - \sin \beta)$.Using trigonometric identities,$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = a \cdot 2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$.This simplifies to $\cot \frac{\alpha-\beta}{2} = a$,so $\alpha - \beta = 2 \cot^{-1} a$.Substituting back,$\sin^{-1} x - \sin^{-1} y = 2 \cot^{-1} a$.Differentiating with respect to $x$,we get $\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-y^2}} \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$.Squaring both sides,$\left(\frac{dy}{dx}ight)^2 = \frac{1-y^2}{1-x^2}$,so $(1-x^2) \left(\frac{dy}{dx}ight)^2 = 1-y^2$.Differentiating again with respect to $x$,we get $(1-x^2) \cdot 2 \frac{dy}{dx} \frac{d^2y}{dx^2} + (-2x) \left(\frac{dy}{dx}ight)^2 = -2y \frac{dy}{dx}$.Dividing by $2 \frac{dy}{dx}$ (assuming $\frac{dy}{dx} 
eq 0$),we get $(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -y$.Multiplying by $(1-x^2)$,we get $(1-x^2)^2 \frac{d^2y}{dx^2} - x(1-x^2) \frac{dy}{dx} = -y(1-x^2)$.Rearranging,$(1-x^2)^2 \frac{d^2y}{dx^2} + y(1-x^2) = x(1-x^2) \frac{dy}{dx}$.Multiplying by $\frac{dy}{dx}$,we get $\left[(1-x^2)^2 \frac{d^2y}{dx^2} + y(1-x^2)ight] \frac{dy}{dx} = x(1-x^2) \left(\frac{dy}{dx}ight)^2$.Since $\left(\frac{dy}{dx}ight)^2 = \frac{1-y^2}{1-x^2}$,the expression becomes $x(1-x^2) \cdot \frac{1-y^2}{1-x^2} = x(1-y^2)$.

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$,then $\left[\left(1-x^2\right)^2 \frac{d^2 y}{d x^2}+y\left(1-x^2\right)\right] \frac{d y}{d x}=$

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