
Suneil P. answered 06/30/14
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No need integrals :)
The tangent function comes from what you said: opp/adj
Imagine a unit circle on a cartesian plane. Now, let us suppose we are working with angles in Quadrant I; suppose we are working with some angle Θ; then we can form a right triangle corresponding to this angle (its three segs are the segment connecting the origin to a point on the circle that is Θ counterclockwise with respect to the x-axis, and then the segment drawn from that point on the circle perpendicularly down to the x-axis, and finally the segment from the origin to this point on the x-axis). The first of the three segs is along what we call the terminal ray and forms the hypotenuse of the said triangle. The segment along the x-axis is part of what we refer to as the initial ray.
Then tan(Θ) is simply opp/adj or the length of the side opposite the angle divided by the length of the side adj to it. Graphically, this is just the y-coordinate divided by the x-coordinate of the terminal point (the intersection of the hypotenuse and the circle). This is just the slope of the hypotenuse.
To graph the function, we plot the tan values for different input angle values of theta.
For a nice illustration of how the tan function is formed, this may help:"
http://www.math.tamu.edu/~yasskin/calclab/DEMOS/trigdef/tan.html

Suneil P.
06/30/14