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Mathematics Question on Continuity and differentiability

If 1−x2+1−y2=x−y\sqrt{1-x^{2} } + \sqrt{1- y^{2}} =x -y , then dydx=\frac{dy}{dx} =

A

−1−y21−x2-\sqrt{\frac{1-y^{2}}{1- x^{2}}}

B

1−x21−y2\sqrt{\frac{1- x^{2}}{1- y^{2}}}

C

−1−x21−y2-\sqrt{\frac{1-x^{2}}{1- y^{2}}}

D

1−y21−x2\sqrt{\frac{1-y^{2}}{1- x^{2}}}

Answer

1−y21−x2\sqrt{\frac{1-y^{2}}{1- x^{2}}}

Explanation

Solution

Put x=sin⁡θ,y=sin⁡θ,x = \sin \theta , y = \sin \theta , we get 1−sin⁡2θ+1−sin⁡2ϕ=sin⁡θ−sin⁡ϕ\sqrt{1- \sin^{2} \theta } + \sqrt{1 -\sin^{2} \phi } =\sin \theta -\sin \phi
⇒cos⁡θ+cos⁡ϕ=sin⁡θ−sin⁡ϕ\Rightarrow \cos\theta +\cos \phi=\sin\theta-\sin\phi
⇒2cos⁡(θ+ϕ2)cos⁡(θ−ϕ2)=2cos⁡(θ+ϕ2)sin⁡(θ−ϕ2)\Rightarrow 2\cos \left(\frac{\theta+\phi}{2}\right) \cos \left(\frac{\theta-\phi}{2}\right) = 2 \cos \left(\frac{\theta +\phi }{2}\right) \sin \left(\frac{\theta -\phi }{2}\right)
⇒tan⁡(θ−ϕ2)=1⇒θ−ϕ2=tan⁡−1(1)=π4\Rightarrow \tan\left(\frac{\theta-\phi}{2}\right) =1\Rightarrow \frac{\theta-\phi}{2} =\tan^{-1} \left(1\right) = \frac{\pi}{4}
⇒θ−ϕ=π2\Rightarrow\theta-\phi=\frac{\pi}{2}
⇒sin⁡−1x−sin⁡−1y=π2       \Rightarrow \sin^{-1} x -\sin^{-1} y = \frac{\pi}{2} \, \, \, \, \, \, \, ...(i)
Differentiating (i) w.r.t. 'x', we get
11−x2−11−y2dydx=0⇒dydx=1−y21−x2\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}}}

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