Let $A = \tan^{-1}(\sqrt{2})$ and assume $\frac{A}{\pi} = \frac{m}{n}$ with integers $m, n$. This gives $nA = m\pi$ and $\tan(A) = \sqrt{2}$. The strategy is to study the sequence $\tan(kA)$ for $k = 1, 2, 3, \dots$. We first observe that $nA = m\pi$ implies that this sequence contains only finitely many different values, namely $\tan(A), \tan(2A), \dots, \tan(nA)$. The reason is the periodicity $\tan(x + \pi) = \tan(x)$. The required contradiction is now achieved by using $\tan(A) = \sqrt{2}$ and the double angle formula
$$\tan(2B) = \frac{2 \tan(B)}{1 - \tan^2(B)}$$
If $\tan(B) = \frac{a}{b}\sqrt{2}$, where $a, b$ are co-prime integers and $a$ is even, then
$$\tan(2B) = \frac{2ab}{a^2 - 2b^2} \sqrt{2}$$
and $2ab$ is even, $b^2 - 2a^2$ is odd. Moreover, $\gcd(ab, b^2 - 2a^2) = 1$, because a prime $p$ which divides $ab$ and $b^2 - 2a^2$ has either to divide $a$ and so also $b$ or has to divide $b$ (which is odd) and $2a^2$, so it divides $a$. But $\gcd(a, b) = 1$ implies that no such prime $p$ can exist. Therefore, we again obtained a fraction in lowest terms and with even numerator. As we have $|2ab| > |a|$ the numerator of $\frac{1}{\sqrt{2}}\tan(2B)$ is bigger than the numerator of $\frac{1}{\sqrt{2}}\tan(B)$.
Because $\frac{1}{\sqrt{2}}\tan(2A) = -2$ has even numerator, we can conclude that in the sequence $\tan(2A), \tan(4A), \tan(8A), \dots, \tan(2^k A), \dots$ no two numbers are equal in contradiction to the finiteness shown above. This contradiction proves that $\frac{A}{\pi}$ cannot be rational.