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

 

Put $x=\tan \theta$ and $y=\tan \varphi$ Then $\sqrt{1+x^2}$

$=\sec \theta ,\sqrt{1+y^2}=\sec \varphi$, and the equation becomes

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

$\begin{array}{l}\Rightarrow \frac{\cos \varphi +\cos \theta }{\cos \theta \cos \varphi }=A\left(\frac{\sin \theta -\sin \varphi }{\cos \theta \cos \varphi }\right)\\ \Rightarrow \cos \varphi +\cos \theta =A(\sin \theta -\sin \varphi )\\ \Rightarrow 2\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=2A\sin \frac{\theta -\varphi }{2}\cos \frac{\theta +\varphi }{2}\end{array}$

$\begin{array}{l}\Rightarrow \cot \frac{\theta -\varphi }{2}=A\Rightarrow \theta -\varphi =2{\cot }^{-1} A\\ \Rightarrow {\tan }^{-1} x-{\tan }^{-1} y=2{\cot }^{-1} A.\end{array}$

Differentiating this, we get $\frac{1}{1+x^2}-\left(\frac{1}{1+y^2}\right)\frac{dy}{dx}=0$,

which is a differential equation of degree 1.