## Precalculus (6th Edition) Blitzer

The required value is $\sqrt{2}-1$
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 $\sqrt{2}-1$