<s>[INST] Calculate $ \arctan(x_{1}) \cdot \arctan(x_{2}) $ 
Let $ x_{1} $ and $ x_{2} $ be the roots of the equation : $ x^2-2\sqrt{2}x+1=0 $
Calculate $ \arctan(x_{1}) \cdot  \arctan(x_{2}) $

The answer should be $ \dfrac{3\pi^2}{64} $.
How does the fact that $ x_{1} $ = $ 1 + \sqrt2 $ = $ \dfrac{1}{x_{2}} $ help?
 [/INST] We have $$x^2-2\sqrt{2}x+1=(x-(1+\sqrt2))(x-(-1+\sqrt2))=0$$ so let $x_1=1+\sqrt2$ and $x_2=-1+\sqrt2$.
Let $a=\tan^{-1}(1+\sqrt2)$. Using the double angle tangent formula, $$\begin{align}\tan2a=\frac{2\tan a}{1-\tan^2a}&\implies\tan2a=\frac{2(1+\sqrt2)}{1-(1+\sqrt2)^2}=-1\end{align}$$ and hence taking the principal angle gives $$\tan^{-1}(1+\sqrt2)=\frac38\pi$$
Let $b=\tan^{-1}(-1+\sqrt2)$. Using the double angle tangent formula, $$\begin{align}\tan2b=\frac{2\tan b}{1-\tan^2b}&\implies\tan2b=\frac{2(-1+\sqrt2)}{1-(-1+\sqrt2)^2}=1\end{align}$$ and hence taking the principal angle gives $$\tan^{-1}(-1+\sqrt2)=\frac18\pi$$ Therefore $$\boxed{\tan^{-1}(x_1)\cdot\tan^{-1}(x_2)=\frac38\pi\cdot\frac18\pi=\frac3{64}\pi^2}$$ as desired.
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