How am I supposed to know that?

You can obtain the correct answer by solving the equation $\sin x = \cos x$ with and without the tangent half angle substitution, which says that $\sin x = \dfrac{2\tan \dfrac{x}{2}}{1+ \tan^{2} \dfrac{x}{2}}$ and $\cos x = \dfrac{1- \tan^{2} \dfrac{x}{2}}{1+\tan^{2} \dfrac{x}{2}}$ .

This isn't really a solution as such but a way to get the correct answer:

$\tan{\dfrac{\pi}{4}}=1 \Rightarrow \arctan{1}=\dfrac{\pi}{4}$

This means we can rule out all the solutions of the form $\arctan{k}$ .

$2 \arctan{k} =\dfrac{\pi} {4} \Rightarrow k=\tan{\dfrac{\pi} {8}}$

$-\pi<\dfrac{\pi} {8}<\dfrac{\pi}{4} \Rightarrow \tan{\dfrac{\pi}{8}}<\tan{\dfrac{\pi}{4}} \Rightarrow k<1$

Clearly $\sqrt{2}+1>1$ so we only have one possible answer remaining so it must be correct:

$\dfrac{\pi}{4}=2 \arctan{(\sqrt{2}-1) }$

$\tan^{2}(x) = \dfrac{1-\cos(2x)}{1+\cos(2x)}$
Put $x = \dfrac{\pi}{8}$
$\tan^{2}\left(\dfrac{\pi}{8}\right)= \dfrac{\sqrt{2}-1}{\sqrt{2}+1} = (\sqrt{2}-1)^{2}$
$\therefore |\tan\left(\dfrac{\pi}{8}\right)| =\sqrt{2} -1$
Since, x is in the first quadrant, $\tan(x) \ge 0$
$\therefore \tan\left(\dfrac{\pi}{8}\right) =\sqrt{2}-1 \rightarrow \dfrac{\pi}{8} = \arctan(\sqrt{2}-1)$
$\dfrac{\pi}{4} = 2\arctan(\sqrt{2}-1)$