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If $y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2$, then $\left(1-x^2\right) y_2-x y_1=$
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The correct answer is:
$4$
$\begin{aligned} & y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2 \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 \sin ^{-1} x}{\sqrt{1-x^2}}-\frac{2 \cos ^{-1} x}{\sqrt{1-x^2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2\left(\sin ^{-1} x-\cos ^{-1} x\right)}{\sqrt{1-x^2}} \\ & \Rightarrow \sqrt{1-x^2} \frac{\mathrm{d} y}{\mathrm{~d} x}=2\left(\sin ^{-1} x-\cos ^{-1} x\right)\end{aligned}$
Differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
& \quad \sqrt{1-x^2} \cdot \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x) \\
& =2\left(\frac{1}{\sqrt{1-x^2}}-\frac{(-1)}{\sqrt{1-x^2}}\right)=\frac{4}{\sqrt{1-x^2}} \\
& \therefore \quad\left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-x \frac{\mathrm{d} y}{\mathrm{~d} x}=4
\end{aligned}
$$
Differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
& \quad \sqrt{1-x^2} \cdot \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x) \\
& =2\left(\frac{1}{\sqrt{1-x^2}}-\frac{(-1)}{\sqrt{1-x^2}}\right)=\frac{4}{\sqrt{1-x^2}} \\
& \therefore \quad\left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-x \frac{\mathrm{d} y}{\mathrm{~d} x}=4
\end{aligned}
$$
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