If  \sqrt{1-x^{2}}+\sqrt{1-y^{2}}= a\left ( x-y \right ),\left | x \right |< 1,\left | y \right |< 1,  show that  \frac{dy}{dx}= \sqrt{\frac{1-y^{2}}{1-x^{2}}}\cdot

 

 

 

 
 
 
 
 

Answers (1)

put x= \sin \theta \; \; \; \; y= \sin \beta
\therefore \sqrt{1-x^{2}}+\sqrt{1-y^{2}}= a\left ( x-y \right )
\Rightarrow \sqrt{1-\sin ^{2}\theta }+\sqrt{1-\sin ^{2}\beta }= a\left ( \sin \theta -\sin \beta \right )
\Rightarrow \cos \theta +\cos \beta = 2a\, \cos \frac{\theta +\beta }{2}\sin \frac{\theta -\beta }{2}
\Rightarrow 2\cos \frac{\theta +\beta }{2}\cos \frac{\theta -\beta }{2}= 2a\, \cos \frac{\theta +\beta }{2}\sin \frac{\theta -\beta }{2}
\Rightarrow \cot \frac{\theta -\beta }{2}= a\; \; \Rightarrow \theta -\beta = 2\cot ^{-1}a
\therefore \sin^{-1}x-\sin^{-1}y= 2\cot ^{-1}a
\Rightarrow \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}}\times \frac{dy}{dx}= 0
\therefore \frac{dy}{dx}= \sqrt{\frac{1-y^{2}}{1-x^{2}}}

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