The best answer to this question depends on the definitions you're using for the trigonometric functions:
Unit circle:
#t# correspond to point #(x,y)# on the circle #x^2+y^2 =1#
Define:
#sint = y#, , #cos t = x#, , #tant = y/x#
The result is immediate.
Point #(x,y)# on the terminal side of #t# and #r=sqrt(x^2+y^2)#
#sint = y/r#, , #cos t = x/r#, , #tant = y/x#
#sint/cost = (y/r)/(x/r)= (y/r)*(r/x) = y/x = tan t#
Right triangle
#sin theta = "opposite"/"hypotenuse" = "opp"/"hyp"#
#cos theta = "adjacent"/"hypotenuse" = "adj"/"hyp"#
#sin theta = "opposite"/"adjacent" = "opp"/"adj"#
The result follow from :
#sin theta / cos theta = ("opp"/"hyp")/("adj"/"hyp") = ("opp"/"hyp")*("hyp"/"adj") = "opp"/"adj" = tan theta #
