Answer
The required value is $\sqrt{2}-1$
Work Step by Step
The given expression $\tan {{22.5}^{\circ }}$ can be expressed in terms of $\tan \frac{{{45}^{\circ }}}{2}$.
$\tan {{22.5}^{\circ }}=\tan \frac{{{45}^{\circ }}}{2}$
Now, it can be further simplified by using the half angle formula $\tan \frac{\alpha }{2}=\frac{\sin \alpha }{1+\cos \alpha }$.
$\begin{align}
& \tan \frac{{{45}^{\circ }}}{2}=\frac{\sin \frac{{{45}^{\circ }}}{2}}{1+\cos \frac{{{45}^{\circ }}}{2}} \\
& =\frac{\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}} \\
& =\frac{\sqrt{2}}{2+\sqrt{2}}
\end{align}$
Then, rationalizing the expression by multiplying and dividing the expression by $\frac{\sqrt{2}}{2-\sqrt{2}}$, we get:
$\begin{align}
& \frac{\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}}{2+\sqrt{2}}.\frac{2-\sqrt{2}}{2-\sqrt{2}} \\
& =\frac{2\sqrt{2}-2}{4-2} \\
& =\frac{2\sqrt{2}-2}{2} \\
& =\sqrt{2}-1
\end{align}$
Hence, the required value is $\sqrt{2}-1$
