Question 573351
If csc of theta is 3 and cos of theta is less than 0, 
----
From that you can say that theta is in Q2 where x is negative and y is positive.
csc = r/y = 3/1 implies that r = 4 and y = 1
---
Solve for "x":
r^2 = x^2 + y^2
4^2 = x^2 + 1
x^2 = 15
x = -sqrt(15) (because x is negative in Q2)
-------
find sin = y/r = 1/3
cos = x/r = -sqrt(15)/4
tan = y/x = -1/sqrt(15)
cot = x/y = -sqrt(15)
sec = r/x = -4/sqrt(15)
=================================
Cheers,
Stan H.