Question

If $\sqrt{1-x^2}+\sqrt{1-y^2}=x-y$, then $\frac{dy}{dx}=$

• A. $-\sqrt{\frac{1-y^2}{1-x^2}}$
• B. $\sqrt{\frac{1-x^2}{1-y^2}}$
• C. $-\sqrt{\frac{1-x^2}{1-y^2}}$
• D. $\sqrt{\frac{1-y^2}{1-x^2}}$

Answer 1

Answer: (D)

Put $x=\sin \theta ,y=\sin \theta ,$ we get $\sqrt{1-{\sin }^2\theta }+\sqrt{1-{\sin }^2\varphi }=\sin \theta -\sin \varphi$
$\Rightarrow \cos \theta +\cos \varphi =\sin \theta -\sin \varphi$
$\Rightarrow 2\cos \left(\frac{\theta +\varphi }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)=2\cos \left(\frac{\theta +\varphi }{2}\right)\sin \left(\frac{\theta -\varphi }{2}\right)$
$\Rightarrow \tan \left(\frac{\theta -\varphi }{2}\right)=1\Rightarrow \frac{\theta -\varphi }{2}={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$
$\Rightarrow \theta -\varphi =\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}x-{\sin }^{-1}y=\frac{\pi }{2}$ ...(i)
Differentiating (i) w.r.t. 'x', we get
$\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$