Well, \(\displaystyle{\tan{{\left({\frac{{\pi}}{{{3}}}}\right)}}}={\frac{{{\sin{{\left({\frac{{\pi}}{{{3}}}}\right)}}}}}{{{\cos{{\left({\frac{{\pi}}{{{3}}}}\right)}}}}}}={\frac{{{\frac{{\sqrt{{{3}}}}}{{{2}}}}}}{{{\frac{{{1}}}{{{2}}}}}}}={\frac{{\sqrt{{{3}}}}}{{{2}}}}{\left({\frac{{{2}}}{{{1}}}}\right)}=\sqrt{{{3}}}\)

However, you can find it in another way:

Just think of it as a \(\displaystyle{\tan{{60}}}°\) and then draw a \(\displaystyle{30}°-{60}°-{90}°\) tringle

And \(\displaystyle{\tan{{60}}}°\) will be equal to \(\displaystyle{\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{a}{d}{j}{a}{c}{e}{n}{t}}}}\) in reference to \(\displaystyle{60}°\) angle. Thus, \(\displaystyle{o}{p}{p}{o}{s}{i}{t}{e}=\sqrt{{{3}}}\) and \(\displaystyle{a}{d}{j}{a}{c}{e}{n}{t}={1}\). Hence,

\(\displaystyle{\tan{{60}}}°={\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{a}{d}{j}{a}{c}{e}{n}{t}}}}={\frac{{\sqrt{{{3}}}}}{{{1}}}}=\sqrt{{{3}}}\)

However, you can find it in another way:

Just think of it as a \(\displaystyle{\tan{{60}}}°\) and then draw a \(\displaystyle{30}°-{60}°-{90}°\) tringle

And \(\displaystyle{\tan{{60}}}°\) will be equal to \(\displaystyle{\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{a}{d}{j}{a}{c}{e}{n}{t}}}}\) in reference to \(\displaystyle{60}°\) angle. Thus, \(\displaystyle{o}{p}{p}{o}{s}{i}{t}{e}=\sqrt{{{3}}}\) and \(\displaystyle{a}{d}{j}{a}{c}{e}{n}{t}={1}\). Hence,

\(\displaystyle{\tan{{60}}}°={\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{a}{d}{j}{a}{c}{e}{n}{t}}}}={\frac{{\sqrt{{{3}}}}}{{{1}}}}=\sqrt{{{3}}}\)