How is $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{2}-1$?

Question

I wondered, why this: $$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$$ is equal to $\sqrt{2}-1$.
Can anyone explain me, why this is equal? :/

Answer 1

\begin{align} \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} &= \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\cdot\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}-1}}\\ &=\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{2-1}}\\ &=\sqrt{\left(\sqrt{2}-1\right)^2}\\\\ &=\sqrt{2}-1 \end{align}

Answer 2

For $a,b>0$, if $ab=1$, then $$ \frac{a}{b}=a^2 $$ and thus $$ \sqrt{\frac{a}{b}}=a. $$

Now, consider $a=\sqrt{2}-1$ and $b=\sqrt{2}+1$.

Answer 3

Since I've expanded on my initial "joke comment", I might as well make it a full joke answer :) By that I mean, nobody in their right mind would take this approach to actually verify that the two quantities are equal: instead, what follows is a good, but limited, way to produce expressions with radicals that look different, but are really not.

The trigonometric function $\tan$ (tangent) has a wide variety of half-angle formulas. I would like to use the following two:

\begin{align*} \tan \frac\theta 2 &= \frac{1 - \sin \theta}{\cos \theta} \tag{1}\\[10pt] \tan \frac\theta 2 &= \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \tag{2} \end{align*}

Evaluating the first at $\theta = \frac{\pi}{4}$, we have that

\begin{align*} \tan \frac{\pi}{8} = \tan \frac{\pi/4}{2} &= \frac{1 - \sin(\pi/4)}{\cos(\pi/4)} \\[7pt] &= \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} \\[7pt] &= \left(1 - \frac{1}{\sqrt{2}}\right) \cdot \frac{\sqrt{2}}{1} \\[5pt] &= \sqrt{2} - 1 \end{align*}

Now, using the second identity with $\theta = \pi/4$, we have

\begin{align*} \tan \frac{\pi}{8} = \tan \frac{\pi/4}{2} &= \sqrt{\frac{1 - \cos(\pi/4)}{1 + \cos(\pi/4)}} \\[7pt] &=\sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}} \\[7pt] &=\sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} \cdot \left(\frac{\sqrt 2}{\sqrt 2}\right)} \\[7pt] &= \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} \end{align*}