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

We have,
$\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a(x - y)$

On putting $x = \sin \alpha$ and $y = \sin \beta ,$ we get
$\sqrt {1 - {{\sin }^2}\alpha } + \sqrt {1 - {{\sin }^2}\beta } = a(\sin \alpha - \sin \beta )$

$\Rightarrow$ $\cos \alpha + \cos \beta = a(\sin \alpha - \sin \beta )$

$\Rightarrow$ $2\cos \frac{{\alpha + \beta }}{2} \cdot \cos \frac{{\alpha - \beta }}{2} = a\left( {2\cos \frac{{\alpha + \beta }}{2} \cdot \sin \frac{{\alpha - \beta }}{2}} \right)$

$\Rightarrow$ $\cos \frac{{\alpha - \beta }}{2} = a\sin \frac{{\alpha - \beta }}{2}$

$\Rightarrow$ $\cot \frac{{\alpha - \beta }}{2} = a$

$\Rightarrow$ $\frac{{\alpha - \beta }}{2} = {\cot ^{ - 1}}a$
$\Rightarrow$ $\alpha - \beta = 2{\cot ^{ - 1}}a$

$\Rightarrow$ ${\sin ^{ - 1}}x - {\sin ^{ - 1}}y = 2{\cot ^{ - 1}}a$ and $y = \sin \beta ]$

On differentiating both sides w.r.t. $x$, we get
$\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{\sqrt {1 - {y^2}} }}\frac{{dy}}{{dx}} = 0$

therefore,$\frac{{dy}}{{dx}} = \frac{{\sqrt {1 - {y^2}} }}{{\sqrt {1 - {x^2}} }} = \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}}$

Hence proved.