Dễ chứng minh:
$ \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2}.\cot
\frac{B}{2}.\cot \frac{C}{2}(1)$ Do đó:
$
\frac{1}{2}\left( {\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} + \cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}} \right) $
$= \frac{1}{2}\left( {\tan \frac{A}{2} + \tan \frac{B}{2} + \tan
\frac{C}{2} + \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}} \right)\\
= \frac{1}{2}\left( {\tan \frac{A}{2} + \cot \frac{A}{2} + \tan \frac{B}{2} + \cot \frac{B}{2} + \tan
\frac{C}{2} + \cot \frac{C}{2}} \right)
$
$\begin{array}{l}
= \frac{1}{2}\left( {\frac{1}{{c{\rm{os}}\frac{A}{2}\sin \frac{A}{2}}} +
\frac{1}{{c{\rm{os}}\frac{B}{2}\sin \frac{B}{2}}} + \frac{1}{{c{\rm{os}}\frac{C}{2}\sin
\frac{C}{2}}}} \right)\\
= \frac{1}{2}\left( {\frac{2}{{\sin A}} + \frac{2}{{\sin B}} + \frac{2}{{\sin C}}} \right) =
\frac{1}{{\sin A}} + \frac{1}{{\sin B}} + \frac{1}{{\sin C}}
\end{array}$
Chứng minh
(1):
$\cot(\frac{A}{2}+\frac{B}{2})=\frac{\cot\frac{A}{2}\cot\frac{B}{2}-1}{\cot\frac{A}{2}+\cot\frac{B}{2}}$
$\Leftrightarrow
\frac{1}{\cot\frac{C}{2}}=\frac{\cot\frac{A}{2}\cot\frac{B}{2}-1}{\cot\frac{A}{2}+\cot\frac{B}{2}}$
$\Rightarrow $ (1)