The degree of the differential equation satisfying $\sqrt{1+x^2}+\sqrt{1+y^2}=A\left(x \sqrt{1+y^2}-y \sqrt{1+x^2}\right)$ is

None of these

Putting $x=\tan \theta$ and $y=\tan \phi$. Then equation becomes

$\sec \theta+\sec \phi=A(\tan \theta \sec \phi-\tan \phi \sec \theta)$

$\Rightarrow \cos \varphi+\cos \theta=A(\sin \theta-\sin \varphi) \Rightarrow 2 \cos \frac{\theta+\phi}{2} \cos \frac{\theta-\phi}{2}=2 A \cos \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2}$

$\Rightarrow \cot \frac{\theta+\phi}{2}=A \Rightarrow \theta-\varphi=2 \cot ^{-1} A \Rightarrow \tan ^{-1} x-\tan ^{-1} y=2 \cot ^{-1} A$

Differentiating, we get $\frac{1}{1+x^2}-\frac{1}{1+y^2} \frac{d y}{d x}=0$

Which is a differential equation of degree 1.

Hence (4) is the correct answer.