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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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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.
\( \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dy}=0\)
\( \frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}\)

 

answered Mar 9, 2013 by thanvigandhi_1
edited Mar 25, 2013 by thanvigandhi_1
 

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