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Every point on the unit circle gives you cosine as the xx-coordinate and sine as the yy-coordinate. But how do you find the other four trigonometric functions without memorizing four new circles?
It turns out you already have all the information you need: tangent, secant, cosecant, and cotangent are built directly from sine and cosine.

The Four Built-in Ratios

In a right triangle with hypotenuse 11, the side opposite θ\theta is sin⁑θ\sin\theta and the adjacent side is cos⁑θ\cos\theta. Because tangent is opposite over adjacent, tangent is defined as the quotient tan⁑θ=sin⁑θcos⁑θ\tan\theta = \frac{\sin\theta}{\cos\theta} (or yx\frac{y}{x}).
The remaining three functions are reciprocal identitiesβ€”flipping a fraction upside down: cosecant is csc⁑θ=1sin⁑θ\csc\theta = \frac{1}{\sin\theta}, secant is sec⁑θ=1cos⁑θ\sec\theta = \frac{1}{\cos\theta}, and cotangent is cot⁑θ=cos⁑θsin⁑θ=1tan⁑θ\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{1}{\tan\theta}.
In the 10th century, Persian astronomer Abu al-Wafa introduced the secant and cosecant functions specifically to simplify shadow-length calculations for sundials and astronomy tables.
How do we put these definitions to work when evaluating an angle on the unit circle?

Evaluating in Any Quadrant

Let's evaluate all four functions for ΞΈ=5Ο€6\theta = \frac{5\pi}{6}. On the unit circle, this angle lands in Quadrant II with coordinates (x,y)=(βˆ’32,12)(x, y) = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right).
First, find tan⁑(5Ο€6)=yx=1/2βˆ’3/2=βˆ’13=βˆ’33\tan\left(\frac{5\pi}{6}\right) = \frac{y}{x} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}, where we rationalize the denominator by multiplying top and bottom by 3\sqrt{3}.
Next, compute the reciprocals: csc⁑(5Ο€6)=1y=11/2=2\csc\left(\frac{5\pi}{6}\right) = \frac{1}{y} = \frac{1}{1/2} = 2, and sec⁑(5Ο€6)=1x=1βˆ’3/2=βˆ’23=βˆ’233\sec\left(\frac{5\pi}{6}\right) = \frac{1}{x} = \frac{1}{-\sqrt{3}/2} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}.
Finally, cot⁑(5Ο€6)=xy=βˆ’3/21/2=βˆ’3\cot\left(\frac{5\pi}{6}\right) = \frac{x}{y} = \frac{-\sqrt{3}/2}{1/2} = -\sqrt{3}.