# If csc z = \frac{17}{8} and cos z= - \frac{15}{17}, then how do you find cot z?

Dec 21, 2014

I would like to note that referring to a right triangle is not always a good idea in trigonometry. In this case, for example, $\cos \left(z\right)$ is negative and, therefore, angle $\angle z$ cannot be an angle in the right triangle.

A much better approach to trigonometric functions is to use a unit circle - a circle of a radius $1$ with a center at the origin of coordinates.
Any point $A$ on a unit circle defines an angle $\angle \alpha$ from the positive direction of the X-axis counterclockwise to a radius from the origin of coordinates to a point $A$.
The abscissa (X-coordinate) of point $A$ is a definition of a function $\sin \left(\alpha\right)$.
The ordinate (Y-coordinate) of point $A$ is a definition of a function $\cos \left(\alpha\right)$.

Then $\tan \left(\alpha\right)$ is, by definition, a ratio $\sin \frac{\alpha}{\cos} \left(\alpha\right)$.
Similarly, by definition,
$\cot \left(\alpha\right) = \cos \frac{\alpha}{\sin} \left(\alpha\right)$
$\sec \left(\alpha\right) = \frac{1}{\cos} \left(\alpha\right)$
$\csc \left(\alpha\right) = \frac{1}{\sin} \left(\alpha\right)$

Using these definitions, from
$\csc \left(z\right) = \frac{1}{\sin} \left(z\right) = \frac{17}{8}$
we can determine
$\sin \left(z\right) = \frac{8}{17}$.
Then, knowing $\sin \left(z\right) = \frac{8}{17}$ and $\cos \left(z\right) = - \frac{15}{17}$ we determine
$\cot \left(z\right) = \cos \frac{z}{\sin} \left(z\right) = \frac{- \frac{15}{17}}{\frac{8}{17}} = - \frac{15}{8}$