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

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A)
Toolbox:
• Put $x = sin \theta\: and \: y = sin\phi$
• Simplify and differentiate after back substituting $\theta\: and \: \phi$
• $cosx+cosy=2cos\large\frac{x+y}{2}.cos\large\frac{x-y}{2}$
• $sinx-siny=2cos\large\frac{x+y}{2}.sin\large\frac{x-y}{2}$
$cos\theta+cos\phi = a(sin\theta-sin\phi)$
$2cos\large\frac{\theta+\phi}{2}\: cos\large\frac{\theta-\phi}{2}=2a\: cos\large\frac{\theta+\phi}{2}\: sin\large\frac{\theta-\phi}{2}$

$\theta-\phi=2tan^{-1}a$
$tan\large\frac{\theta-\phi}{2}=a$

$\Rightarrow 2tan^{-1}a=\theta-\phi$
$\Rightarrow sin^{-1}x-sin^{-1}y=2tan^{-1}a$

Differentiate w.r. to $x$ on both sides to get the result.
$\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dy}=0$
$\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$