**Question:**

If $\mathrm{y} \sqrt{1-\mathrm{x}^{2}}+\mathrm{x} \sqrt{1-\mathrm{y}^{2}}=1$, prove that $\frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\frac{1-\mathrm{y}^{2}}{1-\mathrm{x}^{2}}}$.

**Solution:**

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,

$0=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}$

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