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

1 Answer
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(z)# is negative and, therefore, 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 #/_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(alpha)#.
The ordinate (Y-coordinate) of point #A# is a definition of a function #cos(alpha)#.

Then #tan(alpha)# is, by definition, a ratio #sin(alpha)/cos(alpha)#.
Similarly, by definition,
#cot(alpha)=cos(alpha)/sin(alpha)#
#sec(alpha)=1/cos(alpha)#
#csc(alpha)=1/sin(alpha)#

Using these definitions, from
#csc(z)=1/sin(z)=17/8#
we can determine
#sin(z)=8/17#.
Then, knowing #sin(z)=8/17# and #cos(z)=-15/17# we determine
#cot(z)=cos(z)/sin(z)=(-15/17)/(8/17)=-15/8#