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Following the goniometric properties for $\tan$, $\tan\left(\dfrac{\pi}{3}\right) = \sqrt{3}$.

According to this formula

$$\arctan (x) = y $$ and $$\tan(y) = x$$

If we let $x = \sqrt{3}$ then isn't the only option for $y=\dfrac{\pi}{3}$ ? Or is it just the sign that is the problem in this case?

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    " *goniometric* properties"? And $\;\arctan\sqrt 3=\frac\pi3+k\pi\;,\;\;k\in\Bbb Z\;$ , since $\;\tan\left(\frac\pi3+k\pi\right)=\sqrt3\;$ – 2017-01-18
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    Sorry, I'm not a native English speaker, I don't know the correct terminology. – 2017-01-18
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    $\arctan(x) = y$ where $x=\sqrt{3}$ and $y=\pi / 3$. Where would the equality in the title come from? – 2017-01-18
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    "Goniometric" means "of our relating to the measurement of angles"; it's not the proper English term, but it is English nonetheless. :) – 2017-01-18

2 Answers 2

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$\tan(\pi/3) = \sqrt{3}$. By definition (if we're dealing with real numbers and using the usual "principal value" arctan) $\arctan(\sqrt{3})$ is the unique number $x$ with $-\pi/2 < x < \pi/2$ and $\tan(x) = \sqrt{3}$, and that is $\pi/3$.

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$\tan$ is periodic, so if $\tan(y) = x$ then also $\tan(y + k \pi) = x$ for $k \in \mathbb{Z}$.