Ytukyg

If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.

Where to start?

Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
$$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
for all $a,bin I$ such that $a+bin I$.

Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)

Hint. The functional relation is written also as
$$
f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
$$
So, for $a=b=0$, we have that
$$
f(0)big(1-f^2(0)big)=2f(0)
$$
or
$$
f(0)big(,f^2(0)+1)=0,
$$
and hence $f(0)=0$.

Next, for $b=h$, we have
$$
lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
$$
hence $f$ continuous everywhere.