The solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$, is

$\tan ^{-1} x+\cot ^{-1} x=C$ $\sin ^{-1} x+\sin ^{-1} y=C$ $\sec ^{-1} x+cosec^{-1} x=C$ none of these

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

We have, $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0 \Rightarrow \frac{1}{\sqrt{1-y^2}} d y+\frac{1}{\sqrt{1-x^2}} d x=0$ On integration, we have $\sin ^{-1} y+\sin ^{-1} x=+C$