Show that $r^2 \cot(A/2) \cot(B/2) \cot(C/2) = [ABC]$ In triangle $\Delta~ ABC$, 
$~r~$ is the in-radius and 
$~[ABC]~$ is the area.
Please explain $$ r^2 \cot(A/2) \cot(B/2) \cot(C/2) = [ABC]$$ thanks.
 A: $$\cot(A/2)=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}~ \mbox{etc} ~\mbox{and}~ \Delta=\sqrt{s(s-a)(s-b)(s-c)}.$$ $\Delta$ denotes thr area $[ABC]$ and $s$ is semiparameter. So the given expression (call it $F$), is
$$F=r^2s \frac{\sqrt{s(s-a)(s-b)(s-c)}}{(s-a)(s-b)(s-c)}=\frac{r^2s^2\Delta}{\Delta^2}=\frac{\Delta^3}{\Delta^2}=\Delta.$$
Here $\Delta$ is area and we have used $\Delta=rs.$.
A: 
\begin{align} 
[ABC]&=[IAB]+[IBC]+[ICA]
.
\end{align} 
\begin{align}
[IAB]&=\tfrac12\cdot|AB|\cdot|IC_t|
=\tfrac12(|AC_t|+|BC_t|)\cdot|IC_t|
=\tfrac12(r\cot\tfrac\alpha2+r\cot\tfrac\beta2)\cdot r
=r^2(\tfrac12\,\cot\tfrac\alpha2+\tfrac12\,\cot\tfrac\beta2)
.
\end{align} 
Similarly,
\begin{align}
[IBC]&=
r^2(\tfrac12\,\cot\tfrac\beta2+\tfrac12\,\cot\tfrac\gamma2)
,\\
[ICA]&=
r^2(\tfrac12\,\cot\tfrac\gamma2+\tfrac12\,\cot\tfrac\alpha2)
.
\end{align} 
Hence,
\begin{align} 
[ABC]&=[IAB]+[IBC]+[ICA]
=r^2(\cot\tfrac\alpha2+\cot\tfrac\beta2+\cot\tfrac\gamma2)
.
\end{align} 
And 
\begin{align} 
\cot\tfrac\alpha2+\cot\tfrac\beta2+\cot\tfrac\gamma2
&=
\cot\tfrac\alpha2\cot\tfrac\beta2\cot\tfrac\gamma2
\quad \text{ for }\alpha+\beta+\gamma=180^\circ
\end{align}
is a known triple cotangent identity,
which is easy to check using
substitution
\begin{align} 
\cot\tfrac\gamma2&=
\frac{\cot\tfrac\alpha2+\cot\tfrac\beta2}{\cot\tfrac\alpha2\cot\tfrac\beta2-1}
.
\end{align}
