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If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$, prove that $\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$.


We are given with an equation $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$, 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,

$\sqrt{1-\sin ^{2} A}+\sqrt{1-\sin ^{2} B}=a(\sin A-\sin B)$

$\cos A+\cos B=a(\sin A-\sin B)$

$2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)=a 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

By using $\cos A+\cos B=2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)$ and $\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

$a=\cot \left(\frac{A-B}{2}\right)$

$\cot ^{-1} a=\frac{A-B}{2}$

$2 \cot ^{-1} a=A-B$

$2 \cot ^{-1} a=\sin ^{-1} x-\sin ^{-1} y$

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


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