<s>[INST] How do I justify $\cot 60$ being $\frac{\sqrt{3}}{3}$? Apologies if this is a basic question, my trigonometry skills simply isn't nowhere near up to scratch.
I have a real problem grappling with how $\cot$ functions. At least in my limited understanding, the standard values found in most table simply doesn't reflect how the identity seems to be calculated.
Take for instance $\cot \frac{\pi}{3}$
$\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3}=\frac{1}{2}$
$\tan 60=\frac{\sin 60}{\cos 60}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
I have seen two definitions of $\cot$, $\frac{\cos}{\sin}$ and $\frac{1}{\tan}$. According to both these definitions, the value of $\cot 60$ should be $\frac{1}{\sqrt{3}}$
Instead in most tables the value is claimed to be $\frac{\sqrt{3}}{3}$.
Are my calculations off? Am I reading the table wrong? Or is this actually the same value, written in this way due to some kind of strange convention?
 [/INST] $\ \sqrt{3}\ $ is defined to be the (unique*) positive real number whose square is $\ 3.\ $ We therefore have:
$$ \frac{1}{\sqrt{3}} = 1\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\left( \sqrt{3} \right)^2} = \frac{\sqrt{3}}{3}. $$
$$$$
Alternatively,
$$\frac{1}{\sqrt{3}} = 1\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{9}} = \frac{\sqrt{3}}{3}. $$
*Uniqueness is not usually a part of the definition, but it is not difficult to show that the number is unique...
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