If $\mathrm{y} \sqrt{1-\mathrm{x}^{2}}+\mathrm{x} \sqrt{1-\mathrm{y}^{2}}=1$, prove that $\frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\frac{1-\mathrm{y}^{2}}{1-\mathrm{x}^{2}}}$.
We are given with an equation $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$, we have to prove that $\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$ by using the
given equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,
Put $x=\sin A$ and $y=\sin B$ in the given equation,
$\sin B \sqrt{1-\sin ^{2} A}+\sin A \sqrt{1-\sin ^{2} B}=1$
$\sin B \cos A+\sin A \cos B=1$
$\sin (A+B)=1$
$\sin ^{-1} 1=A+B$
$\frac{\pi}{2}=\sin ^{-1} x+\sin ^{-1} y$
Differentiating we get,
$0=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}$
$\frac{d y}{d x}=\frac{-\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$
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