Answer
See the full explanation below.
Work Step by Step
$\tan \,\left( \frac{\pi }{4}+\alpha \right)-\tan \,\left( \frac{\pi }{4}-\alpha \right)$
By using the identities of trigonometry,
$\tan \,\left( \alpha +\beta \right)=\frac{\tan \,\alpha +\tan \,\beta }{1-\tan \,\alpha \tan \,\beta }$
And,
$\tan \,\left( \alpha -\beta \right)=\frac{\tan \,\alpha -\tan \,\beta }{1+\tan \,\alpha \tan \,\beta }$
Now, the above expression can be further simplified as,
$\begin{align}
& \tan \,\left( \frac{\pi }{4}+\alpha \right)-\tan \,\left( \frac{\pi }{4}-\alpha \right)=\frac{\tan \,\frac{\pi }{4}+\tan \,\alpha }{1-\tan \,\frac{\pi }{4}\tan \,\alpha }-\frac{\tan \,\frac{\pi }{4}-\tan \,\alpha }{1+\tan \,\frac{\pi }{4}\tan \,\alpha } \\
& =\frac{1+\tan \,\alpha }{1-\tan \,\alpha }-\frac{1-\tan \,\alpha }{1+\tan \,\alpha }
\end{align}$
Now, taking LCM of $\left( 1-\tan \alpha \right)$ and $\left( 1+\tan \alpha \right)$ ,
$\begin{align}
& \frac{1+\tan \,\alpha }{1-\tan \,\alpha }-\frac{1-\tan \,\alpha }{1+\tan \,\alpha }=\frac{\left( 1+\tan \,\alpha \right)\left( 1+\tan \,\alpha \right)-\left( 1-\tan \,\alpha \right)\left( 1-\tan \,\alpha \right)}{\left( 1-\tan \,\alpha \right)\left( 1+\tan \,\alpha \right)} \\
& =\frac{\left( 1+\tan \,\alpha +\tan \,\alpha +{{\tan }^{2}}\,\alpha \right)-\left( 1-\tan \,\alpha -\tan \,\alpha +{{\tan }^{2}}\,\alpha \right)}{\left( {{1}^{2}}-{{\tan }^{2}}\,\alpha \right)} \\
& =\frac{\left( 1+2\tan \,\alpha +{{\tan }^{2}}\,\alpha \right)-\left( 1-2\tan \,\alpha +{{\tan }^{2}}\,\alpha \right)}{\left( 1-{{\tan }^{2}}\,\alpha \right)} \\
& =\frac{1+2\tan \,\alpha +{{\tan }^{2}}\,\alpha -1+2\tan \,\alpha -{{\tan }^{2}}\,\alpha }{\left( 1-{{\tan }^{2}}\,\alpha \right)}
\end{align}$
The above expression is simplified further,
$\begin{align}
& \frac{1+2\tan \,\alpha +{{\tan }^{2}}\,\alpha -1+2\tan \,\alpha -{{\tan }^{2}}\,\alpha }{\left( 1-{{\tan }^{2}}\,\alpha \right)}=\frac{2\tan \,\alpha +2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha } \\
& =\frac{4\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha } \\
& =2\frac{2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha }
\end{align}$
Since, $\tan 2\alpha =\frac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }$. This implies,
$2\frac{2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha }=2\tan \,2\alpha $
Thus, the left side of the given expression is equal to the right side, which is,
$\tan \,\left( \frac{\pi }{4}+\alpha \right)-\tan \,\left( \frac{\pi }{4}-\alpha \right)=2\tan \,2\alpha $.
