Does the following trignometric identity hold? I did some algebraic manipulation and arrived at the following statement:
$\tan{\frac{x}{2}}=\frac{\tan{x}}{\sqrt{\tan^2{x}+1}+1}$. I can't find anything about this identity on the web, but it seems to work for a lot of examples I've tried.
Note that this identity isn't a guess. I did some actual work to arrive to it.
For example:
$\tan{30}=\frac{\tan{60}}{\sqrt{\tan^2{60}+1}+1}$ 
$\tan{15}=\frac{\tan{30}}{\sqrt{\tan^2{30}+1}+1}$
It seems to work with all the halves of standard angles, and their subsequent halves. Is this an identity?
 A: Your identity holds whenever $\cos x>0$, because then we have
$$\frac{\tan x}{\sqrt{\tan^2x+1}+1}=\frac{\tan x}{\sqrt\frac1{\cos^2x}+1}=\frac{\sin x}{1+\cos x}=\frac{2\sin\frac x2\cos\frac x2}{2\cos^2\frac x2}=\tan\frac x2.$$
When $\cos x<0$, an analogous computation gives
$$\frac{\tan x}{\sqrt{\tan^2x+1}+1}=\frac{\sin x}{-1+\cos x}=-\cot\frac x2.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Since $\ds{\quad\tan\pars{x} = {2\tan\pars{x/2} \over 1 - \tan^{2}\pars{x/2}}}$:

\begin{align}
&\color{#44f}{%
\tan\pars{x} \over \root{\tan^{2}\pars{x} + 1} + 1}
\\[5mm] = & \
{2\tan\pars{x/2} \over
1 - \tan^{2}\pars{x/2} -
\bracks{1  +\tan^{2}\pars{x/2}}
\on{sgn}\pars{\tan^{2}\pars{x/2} - 1}}
\\[5mm] = & \
\bbx{\color{#44f}{\left\{\begin{array}{rcl}
\ds{\tan\pars{x \over 2}} & \color{black}{\mbox{if}} &
\ds{\tan^{2}\pars{x \over 2} < 1}
\\[2mm]
\ds{\to \infty} & \color{black}{\mbox{if}} &
\ds{\tan^{2}\pars{x \over 2} \to 1^{-}}
\\[2mm]
\ds{\to -\infty} & \color{black}{\mbox{if}} &
\ds{\tan^{2}\pars{x \over 2} \to 1^{+}}
\\[2mm]
\ds{-\cot\pars{x \over 2}} & \color{black}{\mbox{if}} &
\ds{\tan^{2}\pars{x \over 2} > 1}
\end{array}\right.}} \\ &
\end{align}
