The differential equation: $ \arctan (y) = \arctan(x)+C .$ I solved the equation and stalled. Help with decision please.
$$(1+y^2)\,dx=(1+y^2)\,dy \iff \int \frac{dx}{1+x^2} = \int\frac{dy}{1+y^2}  $$ 
Transformed expression for the table of integrals.
$$ \arctan (y) = \arctan(x)+C $$
Prompt how to further transform expression.(Find the general solution)

Take the tangent of each side of the equation$$\tan(\arctan(\theta)) = \theta$$
$$\tan\Big(\arctan(y)\Big) =\tan\Big(\arctan(x) + C\Big) \iff y = \tan\Big(\arctan(x) + C\Big)$$
As Lucian notes, you can use the following double-angle formula for $\tan$ to expand the right-hand side of the result:
$$\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a) \tan(b)}$$
Here, put $a = \arctan(x)$, and $b = C$. 
