Find tan(C/2) in The triangle ABC . $ABC$ is a triangle.
$\tan\frac{A}{2}  = 0.5$
$\tan\frac{B}{2}  =\frac{1}{3}$.
Find $\tan\frac{C}{2}$.
I tried to find it :
when $A+B+C = 180^{\circ}$ 
So $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}$.
How can I complete on these information?

The following identity is very useful.
$$\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{A}{2}\tan\frac{C}{2}+\tan\frac{B}{2}\tan\frac{C}{2}=1$$
From here $C=90^{\circ}$
We can get this result by geometric way.
Let $P(0,0)$,$Q(2,1)$ and $R(3,-1)$. Draw it!
Hence $PQ=QR$, $\measuredangle Q=90^{\circ}$ and from here $$\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}=\measuredangle Q=45^{\circ},$$
which says that our $\frac{C}{2}=45^{\circ}$ and we are done!
