Question
यदि $\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a(x - y)$, तो $\frac{{dy}}{{dx}} = $
$\cos \theta + \cos \phi= a(\sin \theta - \sin \phi)$
==> $2\cos \frac{{\theta + \phi}}{2}\cos \frac{{\theta - \phi}}{2} = a\left\{ {2\cos \frac{{\theta + \phi}}{2}\sin \frac{{\theta - \phi}}{2}} \right\}$
==> $\frac{{\theta - \phi}}{2} = {\cot ^{ - 1}}a \Rightarrow \theta - \phi = 2{\cot ^{ - 1}}a$
==> ${\sin ^{ - 1}}x - {\sin ^{ - 1}}y = 2{\cot ^{ - 1}}a$
==> $\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{\sqrt {1 - {y^2}} }}\frac{{dy}}{{dx}} = 0 $
$\Rightarrow \frac{{dy}}{{dx}} = \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} $.
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