Additional formula for tangent function? I am having some trouble with this one problem.
The question is:
Suppose we wish to find a real-valued, differentiable function $F(x)$ that satisfies the functional
equation $$F(x+y) =\frac{F(x)+F(y)}{1-(F(x)*F(y))}$$
Show that $F$ necessarily satisfies $F(0) = 0$. Hint: Use the above to get an expression for F(0+0) and then use the fact that we seek F to be real-valued.
Set $a = F'(0)$. Show that $F$ must satisfy the diff  $\frac{dF}{dx} = a(1+F(x)^2)$
Hint: Differentiate the above WRT to $y$ then set $y=0$
I don't quite understand how to show that $F(0)=0$ and how to use the fact that $F$ is real valued to solve the problem. Any help would be appreciated, thanks!
 A: Hint: Let $F(0)=x$, then
$$x=F(0+0)=\frac{2x}{1-x^2}.$$
Now, find the real solutions of
$$x=\frac{2x}{1-x^2}.$$
A: First,
$$
F(0+0)=\frac{F(0)+F(0)}{1-F(0)\cdot F(0)}\tag{1}
$$
Assume that $F(0)\ne0$, then divide $(1)$ by $F(0)$ to get
$$
1=\frac2{1-F(0)^2}\tag{2}
$$
which implies $F(0)^2=-1$. Therefore, $F(0)=0$.

Next,
$$
\begin{align}
F'(x)
&=\lim_{h\to0}\frac{F(x+h)-F(x)}h\\
&=\lim_{h\to0}\frac{\frac{F(x)+F(h)}{1-F(x)F(h)}-F(x)}h\\
&=\left(1+F(x)^2\right)\lim_{h\to0}\frac{F(h)}h\\
&=\left(1+F(x)^2\right)F'(0)\tag{3}
\end{align}
$$
That is,
$$
\arctan(F(x))=F'(0)\,x+C\tag{4}
$$
Since $F(0)=0$, we get $C=0$. Thus, we have
$$
F(x)=\tan(ax)\tag{5}
$$
A: I will try to derive
the differential equation
from the functional equation.
Spoiler:
I almost succeed,
but haven't been able
to show that
$f'(0) = 1$.
If
$f(x+y) 
=\frac{f(x)+f(y)}{1-f(x)f(y)}
$,
$\begin{array}\\
f(x+h)-f(x) 
&=\frac{f(x)+f(h)}{1-f(x)f(h)}-f(x)\\
&=\frac{f(x)+f(h)-f(x)+f^2(x)f(h)}{1-f(x)f(h)}\\
&=\frac{f(h)+f^2(x)f(h)}{1-f(x)f(h)}\\
&=f(h)\frac{1+f^2(x)}{1-f(x)f(h)}\\
\end{array}
$
Note that this is where
$1+f^2(x)$ comes in.
Dividing by $h$,
$\frac{f(x+h)-f(x)}{h} 
=\frac{f(h)}{h}\frac{1+f^2(x)}{1-f(x)f(h)}
$.
Letting
$h \to 0$,
since $f(0) = 0$,
$f'(x)
=f'(0)(1+f^2(x))
$.
Also note that,
setting
$y=-x$,
we get
$f(x)+f(-x) = 0$.
The final step 
I would like to do here
is to show that
$f'(0)=1$.
Unfortunately,
I don't see how to do this
right now,
so I'll leave it at this.
Probably someone else
will see what I have overlooked.
A: $$F(0+0)=\frac{2F(0)}{1-F^2(0)}\implies F(0)=0,\pm i$$

$$\newcommand{\d}[1]{\frac{{\rm d}}{{\rm d}x}\left(#1\right)}
\d{ F(x+y)}=\d{\frac{F(x)+F(y)}{1-F(x)F(y)}}=\frac{1+F^2(y)}{(1-F(x)F(y))^2}\\
F'(y)=1+F^2(y)\\\implies F(x)=\tan x+{\cal C },\because F(0)=0\implies {\cal C}=0\\F(x)=\tan x$$

$$\bbox[5pt,border:2pt solid black]{\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}}$$
