The general solution of the differential equation $\frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ is :

$sin^{-1}x+sin^{-1}y=C$ (where C is constant of integration)

The correct answer is Option (2) → $sin^{-1}x+sin^{-1}y=C$ (where C is constant of integration)

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

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

so $\sin^{-1}y=\sin^{-1}x×(-1)+C$

so $\sin^{-1}x+\sin^{-1}y=C$