I'm going to write the simplification out, because it took me a while.
\tan^{-1}y - \tan^{-1}x = C
In order to take tan on the left side, this equation needs to be re-written as:
\tan^{-1}\frac{y-x}{1+yx}=C
then we can take the tangent of both sides, giving us:
\frac{y-x}{1+yx}=C
then we...
Obtain the solution to the differential equation:
\frac{dy}{dx} = \frac{1+y^2}{1+x^2}
Multiple choice answer:
a) \frac{Cx}{1-Cx}
b) \frac{Cx}{1+Cx}
c) \frac{C-x}{1-Cx}
d) \frac{1-Cx}{x+C}
e) \frac{x+C}{1-Cx}
Tried integrating two sides to arrive at arctan y = arctan x + C, but not sure...