Question

$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$

then prove that $\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$

Answer 1

$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$

Let $x=\sin A,y=\sin B$

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

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

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

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

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

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

Substituting the values of $x,y$ we get

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

Differentiate w.r.t $x$, we get

$\Rightarrow \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$

$\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$

Hence Proved.