tan (##pi/3##)==##sqrt3##

tan (##pi/3##)= tan(##(180^o)/3##) = tan(##60^o##)=##sqrt3##

{##NOTE##: I have substituted ##pi^c = 180^o## }

Now, there’s a given way of remembering sine and cosine values of 5 crucial angles, which are, ##0^o##, ##30^o##, ##45^o##, ##60^o##, and ##90^o##.

The sine values: sin(##0^o##) = ##0##

sin(##30^o##) = ##1/2##

sin(##45^o##) = ##1//(sqrt2)##

sin(##60^o##) = ##(sqrt3)//2##

sin(##90^o##) = ##1##

For the cosine values, you need to refer to the same, but in the reverse order. ##rArr## cos(##0^o##) = ##1##

cos(##30^o##) = ##(sqrt3)/2##

cos(##45^o##) = ##1/(sqrt2)##

cos(##60^o##) = ##1/2##

cos(##90^o##) = ##0##

Finally, you can use this relation to find the tangents of any angle, ##tan theta## = ##(sin theta)/(cos theta)## FOR example, ##tan 45^o = (sin 45^o)/(cos 45^o) ## = ##(1/(sqrt2))//(1/(sqrt2))## = ##1##

I hope it helps.