I am told that $x_1$ and $x_2$ are the roots of the following equation:
$$ x^2 - 2\sqrt{2}x + 1 = 0 $$
And I have to find the following:
$$\hspace{6cm} \arctan x_1 + \arctan x_2 \hspace{5cm} (1)$$
$$\hspace{6.cm} \arctan x_1 \cdot \arctan x_2 \hspace{5.3cm} (2)$$
Now I know that, in general:
$$\arctan(a) + \arctan(b) = \arctan \bigg ( \dfrac{a + b}{1 - ab} \bigg )$$
So $(1)$ is not that difficult to find, it actually equals $\dfrac{\pi}{2}$. The problem is that I don't know how to find $(2)$.
With $x_1+x_2=2\sqrt2$ and $x_1x_2=1$,
$$S_+=\arctan x_1 + \arctan x_2 = \arctan \bigg ( \dfrac{x_1 + x_2}{1 - x_1x_2} \bigg )=\arctan (\infty) =\frac\pi2$$
With $x_1-x_2= \sqrt{(x_1+x_2)^2-4x_1x_2}=2$,
$$S_-=\arctan x_1 - \arctan x_2 = \arctan \bigg ( \dfrac{x_1 - x_2}{1 + x_1x_2} \bigg )=\arctan (1)=\frac\pi4$$
Then, $$\arctan x_1\cdot \arctan x_2 = \frac14(S_++S_-)(S_+-S_-) =\frac14(S_+^2-S_-^2)=\frac{3\pi^2}{64}$$
As far as I can tell, there is no clean, generalized way to find $\arctan(x) \cdot \arctan(y)$. However, the roots for this specific equation are $\sqrt{2}-1$ and $\sqrt{2}+1$. $\arctan{\left(\sqrt{2}-1\right)} = \frac{\pi}{8}$ and $\arctan{\left(\sqrt{2}+1\right)} = \frac{3\pi}{8}$. So in this case we can directly calculate the product $\frac{3\pi^2}{64}$.
With $x=\tan a$, $$\tan^2a-2\sqrt2\tan a+1=0$$ can easily be rewritten as $$\sin 2a=\sin\frac\pi4.$$
Thus the solutions with $a\in(-\frac\pi2,\frac\pi2)$ are
$$2a_1=\frac\pi4\text{ and }2a_2=\frac{3\pi}4,$$
giving $$a_1a_2=\frac{3\pi^2}{64}.$$
The roots of the equation are $ x_1 = \sqrt{2} + 1$ and $ x_2 =\sqrt{2} - 1 $. So we can explicitly work out $ \arctan(x_1) \, \cdot \, \arctan(x_2) = \frac{\pi}{8} \, \cdot \, \frac{3\pi}{8} = \frac{3\pi^2}{64} $
