Since the tangent function is related to sin(t) and cos(t) --- tan(t)/1=sin(t)/cos(t) --- and continues to work with the unit circle, we start with the same steps we use with sin(t) and cos(t) to find the point (t,tan(t)). With the equation above we can use the idea of similar triangles. We have the smaller triangle that starts from point C perpendicular to the x-axis (sin(t)) and we have the line across the x-axis (cos(t)), we use the line that x=1 (tan(t) / 1) to create point D, which is (1,tan(t)), the side parallel to sin(t) is tan(t) - Line BD. But now since we want to get to (t, tan(t)) we must use the intersection of the tan(t) triangle to get the intersection to x=t (the curve). Which is why we use point D as the perpendicular line to y-axis. Which then intersects line a, giving us the necessary point E which is (t,tan(t)). This works because since x=t is based on the position of the curve on the circle and there is an intersection point with tangent, it gives us (t,tan(t)). D = (1, tan(t), by the use of cosine and sine functions. E = (t, tan(t)) because it incorporates the line with x=t, which allows the point to be made. C is a point that rotates along our unit circle, giving the arc (t).