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?
 A: 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.
A: $$\csc2t-\cot2t=\tan t$$
Here $2t=2m\pi + \dfrac\pi4$ where $m$ is any integer as $\csc2t =\sqrt2=\csc\dfrac\pi4$ and $\cot2t=1=\cot?$
But $-\dfrac\pi2<\arctan(\tan t)<\dfrac\pi2$ using  https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values
Similarly, 
$$\csc2t+\cot2t=\cot t=\tan\left(\dfrac\pi2-t\right)$$
A: use that $$\arctan(\sqrt{2}+1)=\frac{3}{8}\pi$$
