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

$\sqrt{\frac{1-y^2}{1-x^2}}$

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

let $x=\sin A$ $y=\sin B$

from (1) $\sqrt{1-\sin A^2}+\sqrt{1-\sin B^2}=a(\sin A-\sin B)$

$\cos A+\cos B = a(\sin A-\sin B)$

$=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}=a(2\sin\frac{A-B}{2}\sin\frac{A+B}{2})$

$⇒\cot\frac{A-B}{2}=a$

so $A - B = 2\cot^{-1}a$

$⇒\sin^{-1}x-\sin^{-1}y= 2\cot^{-1}a$

differentiating w.r.t x

$\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$

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