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

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? :/

• Hint: Multiply top and bottom of the inside by $\frac{\sqrt{2}-1}{\sqrt{2}-1}$, which is just a fancy way of multiplying by $1$ and wouldn't change the value. Jan 9, 2017 at 20:10
• Because they're both equal to $\tan\left(\frac\pi 8\right)$ Jan 9, 2017 at 20:13
• @RSerrao Mostly for fun (I wanted to put an emoticon, but it clashed with the math), but to be fair, it was one of the first things that came to mind Jan 9, 2017 at 20:22
• @pjs36 how does one find that $\sqrt{2} - 1 = \tan(\frac\pi8)$ without resorting to a calculator?!
– RGS
Jan 9, 2017 at 20:23
• @RSerrao Tangent has a wide variety of half-angle formulas. I remember this one in particular because I created an answer key for a quiz and tried to anticipate various solution methods. Relevant here is that $\tan \frac\theta 2 = \frac{1 - \cos \theta}{\sin \theta} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$. Use the first for $\sqrt{2} - 1$, and the latter for the nested radicals. Jan 9, 2017 at 20:33

\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}

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 $b=\sqrt{2}+1$.

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*}