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

Question

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!

Answer 1

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

$$\arctan x=\arctan\tan\left(\frac{\pi}{2}-y\right)=\frac{\pi}{2}-y=\frac{\pi}{2}-\arctan \frac1x$$

Answer 2

Using complex numbers:

Let $z = x + y \, i$ and $w = y + x \, i$. Then

$$ \arctan (\frac xy) +\arctan (\frac yx) = \arg w + \arg z = \arg wz = \arg i (x^2 + y^2) = \frac{\pi}{2} $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 4

Funny I played with it too

$$E=\arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)=x_1+x_2$$

$$ \tan(E)=\frac {\tan(x_1)+\tan(x_2)}{1-\tan(x_1)\tan(x_2)}$$

$$\tan(E)=\pm\infty$$

$$ \vdots $$