Question

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a\left(x-y\right)$, prove that $\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{1-x^2}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},\sqrt{1-x^2}+\sqrt{1-y^2}=a\left(x-y\right) \\ \mathrm{Let}x=\sin A,y=\sin B \\ \Rightarrow \sqrt{1-{\sin }^{2}A}+\sqrt{1-{\sin }^{2}B}=a\left(\sin A-\sin B\right) \\ \Rightarrow \cos A+\cos B=a\left(\sin A-\sin B\right) \\ \Rightarrow a=\frac{\cos A+\cos B}{\sin A-\sin B} \\ \Rightarrow a=\frac{2\cos {\displaystyle \frac{A+B}{2}}\cos {\displaystyle \frac{A-B}{2}}}{2\cos {\displaystyle \frac{A+B}{2}}\sin {\displaystyle \frac{A-B}{2}}}\left[\begin{array}{c}\because \sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}\\ \because \cos A+\cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}\end{array}\right] \\ \Rightarrow a=\cot \left(\frac{A-B}{2}\right) \\ \Rightarrow {\cot }^{-1}a=\frac{A-B}{2} \\ \Rightarrow 2{\cot }^{-1}a=A-B \\ \Rightarrow 2{\cot }^{-1}a={\sin }^{-1}x-{\sin }^{-1}y\left[\because x=\sin A,y=\sin B\right] \end{gathered}$$
Differentiating with respect to x, we get,
$$\begin{gathered} \frac{d}{dx}\left(2co{t}^{-1}a\right)=\frac{d}{dx}\left({\sin }^{-1}x\right)-\frac{d}{dx}\left({\sin }^{-1}y\right) \\ \Rightarrow 0=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}} \\ \Rightarrow \frac{1}{\sqrt{1-y^2}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{1}{\sqrt{1-x^2}} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\sqrt{\frac{1-y^2}{1-x^2}} \end{gathered}$$