Solution of $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ is :

$\sin ^{-1} x-\sin ^{-1} y=c$ $\sin ^{-1} y+\sin ^{-1} x=c$ $\sin ^{-1} x=c \sin ^{-1} y$ $\left(\sin ^{-1} x\right)\left(\sin ^{-1} y\right)=c$

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

$\frac{d y}{\sqrt{1-y^2}}=\frac{-d x}{\sqrt{1-x^2}} \Rightarrow \sin ^{-1} y+\sin ^{-1} x=c$ Hence (2) is the correct answer.