Solution
= Solution
Since
$$
\frac{f'}{1+f^2}=C,
$$
integration gives
$$
\arctan f(x)=Cx+D.
$$
Part (i) gives $D=0$, so the <tangent addition functional equation> has the solutions
$$
\boxed{f(x)=\tan(Cx)},
$$
for constants $C$ such that no pole lies in $(-a,a)$. Conversely, the tangent addition formula verifies the functional equation. If $f'(0)=2$, then $C=2$, and
$$
\boxed{f(x)=\tan(2x)}
$$
on any such interval, necessarily with $a\leq\pi/4$.