What is the value of $\arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)?$ I was playing about with some numbers when I came up with this fun question.

What is the value of $\arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)?$

Here is my method:

As is clearly evident from the triangle:
$a = \arctan \left(\frac yx\right)$ and
$b = \arctan \left(\frac xy\right)$
$\therefore \arctan \left(\frac xy\right) +\arctan \left(\frac yx\right) = a + b = 90^{\circ} = \frac {\pi}2 ^c$
Was my method right? Or can it be improved? I would appreciate any help in the comments or through answers. Thanks in advance!
 A: Yes it's a correct method.
As an alternative note that for $x>0$
$$\arctan x + \arctan \frac1x = \frac{\pi}2$$
indeed if you set 
$$y=\arctan \frac1x$$ 
then 
$$\tan y=\frac1x$$ 
that is
$$x=\cot y=\tan\left(\frac{\pi}{2}-y\right)$$
therefore
$$\arctan x=\arctan\tan\left(\frac{\pi}{2}-y\right)=\frac{\pi}{2}-y=\frac{\pi}{2}-\arctan \frac1x$$
A: Using complex numbers:
Let $z = x + y \, i$ and $w = y + x \, i$. Then
$$
\arctan (\frac xy) +\arctan (\frac yx)
= \arg w + \arg z
= \arg wz
= \arg i (x^2 + y^2)
= \frac{\pi}{2}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\arctan\pars{x \over y}\ +\
\overbrace{\quad\qquad\arctan\pars{y \over x}\quad\qquad}
^{\ds{{\pi \over 2}\,\mrm{sgn}\pars{x \over y} - \arctan\pars{x \over y}}} & =
\bbx{{\pi \over 2}\,\mrm{sgn}\pars{x \over y}}
\end{align}
A: Funny I played with it too
$$E=\arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)=x_1+x_2$$
$$ \tan(E)=\frac {\tan(x_1)+\tan(x_2)}{1-\tan(x_1)\tan(x_2)}$$
$$\tan(E)=\pm\infty$$
$$ \vdots $$
