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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:
enter image description here
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!

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  • $\begingroup$ I assure you that both $x$ and $y$ are larger than $0$ else that triangle would have never appeared. $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 14:51
  • $\begingroup$ Ok then your answer is correct according to me! You wrote the expression, triangle was brought by you afterwards $\endgroup$ – samjoe Jan 5 '18 at 14:52
  • $\begingroup$ @samjoe this is the first time I used $\arctan$ so I wanted to ask if it is identical to the $\tan^{-1}$ we all learnt before high school. $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 14:54
  • $\begingroup$ Thanks for the explanation! $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 14:54
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    $\begingroup$ Using \circ for degree symbol looks better in my opinion e.g compare $90^\circ$ and $90^0$ $\endgroup$ – kingW3 Jan 5 '18 at 15:43
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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

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

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  • $\begingroup$ Should I close this question? As this is technically a duplicate as labbhattacharjee's link shows. $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 15:12
  • $\begingroup$ @MohammadZuhairKhan Probably some moderator will set it as a duplicate with a link at the previous. $\endgroup$ – gimusi Jan 5 '18 at 15:14
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    $\begingroup$ Gimusi. Very nice. $\endgroup$ – Peter Szilas Jan 5 '18 at 15:27
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Using complex numbers:

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

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

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  • $\begingroup$ I intended to use $\frac {\pi}2$ but for a triangle isn't $90^0$ a more common convention? $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 15:26
  • $\begingroup$ Why $\arg w + \arg z = \arg wz$? $\endgroup$ – Did Jan 8 '18 at 18:21
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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    $\begingroup$ Could you explain what does $sgn$ mean? $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 15:41
  • $\begingroup$ I guess it means sign ... $\endgroup$ – Isham Jan 5 '18 at 15:49
  • $\begingroup$ @MohammadZuhairKhan $\mathrm{sgn}$ is the $\texttt{sign}$ function. $\mathrm{sgn}:\mathbb{R} \to \mathbb{Z}$. $$ \mathrm{sgn}\left(x\right) \equiv \left\{\begin{array}{rcrcl} {\displaystyle -1} & \mbox{if} & {\displaystyle x} & {\displaystyle <} & {\displaystyle 0} \\ {\displaystyle 0} & \mbox{if} & {\displaystyle x} & {\displaystyle =} & {\displaystyle 0} \\ {\displaystyle 1} & \mbox{if} & {\displaystyle x} & {\displaystyle >} & {\displaystyle 0} \end{array}\right. $$ $\endgroup$ – Felix Marin Jan 5 '18 at 17:23
  • $\begingroup$ @FelixMarin thanks for explaining. I had never seen this function before so my apologies. $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 17:25
  • $\begingroup$ @MohammadZuhairKhan Thanks. You're welcome. Also, it's related to the $\texttt{Heaviside Step Function}$ $\mathrm{H}$: $\mathrm{sgn}\left(x\right) = 2\,\mathrm{H}\left(x\right) - 1$ when $x \not= 0$.. $\endgroup$ – Felix Marin Jan 5 '18 at 17:28
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Funny I played with it too

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

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

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

$$ \vdots $$

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    $\begingroup$ $E = \arctan \infty = $(b.a.a.) $\frac {\pi}2 $ $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 15:59
  • $\begingroup$ yep @MohammadZuhairKhan +or - $\frac {\pi} 2$...I didnt understand why I got a zero at the denominator but it makes sense. $\endgroup$ – Isham Jan 5 '18 at 16:01
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    $\begingroup$ $\tan$ is periodic. I prefer $\pi \ge x \ge 0$ whenever possible. $\endgroup$ – Mohammad Zuhair Khan Jan 5 '18 at 17:20

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