Apply the definition:
#tan(x) = \frac{sin(x)}{cos(x)}#
So,
#tan(-pi) = \frac{sin(-pi)}{cos(-pi)}#
Actually, you may notice that #-pi# and #pi# identify the same angle, since you're making half of a turn, either clockwise or counterclockwise, but still ending at #(-1,0)#. So, we may simplify the expression into
#tan(-pi) = tan(pi) = \frac{sin(pi)}{cos(pi)}#
Now, #pi# is a known value for trigonometric function, and I've actually already wrote the answer: since the angle #pi# is associated to the point #(-1,0)#, and for every point #P=(x,y)# identified by the angle #alpha# on the unit circumference you have #(cos(alpha),sin(alpha))=(x,y)#, we have
#(cos(pi),sin(pi))=(-1,0)#
and thus
#tan(-pi) = tan(pi) = \frac{sin(pi)}{cos(pi)} = \frac{0}{-1}=0#