If y = (sin^{-1} x)^2 + (cos^{-1} x)^2,then ( | vedclass

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(B) Given $y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$.We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,so $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1}

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$4$

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(B) Given $y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$.We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,so $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.Substituting this,$y = (\sin^{-1} x)^2 + (\frac{\pi}{2} - \sin^{-1} x)^2$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} + 2(\frac{\pi}{2} - \sin^{-1} x) \cdot (-\frac{1}{\sqrt{1-x^2}})$.$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} (\sin^{-1} x - \frac{\pi}{2} + \sin^{-1} x) = \frac{2(2\sin^{-1} x - \frac{\pi}{2})}{\sqrt{1-x^2}}$.Multiplying by $\sqrt{1-x^2}$: $\sqrt{1-x^2} y_1 = 4\sin^{-1} x - \pi$.Differentiating again with respect to $x$:$\sqrt{1-x^2} y_2 + y_1 \cdot (\frac{-2x}{2\sqrt{1-x^2}}) = \frac{4}{\sqrt{1-x^2}}$.Multiplying throughout by $\sqrt{1-x^2}$: $(1-x^2) y_2 - x y_1 = 4$.

If $y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$,then $(1 - x^2) y_2 - x y_1 =$

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