SOLUTION: Find the values of the trigonometric functions of t from the given information. csc(t) = 3, cos(t) < 0

3 is a positive number.  The cosecant is positive in QI and QII.
The cosine is negative in QII and QIII.  Therefore t must be in
QII.

The cosecant is r/y or the hypotenuse over the opposite.  Therefore we
take 3 as the fraction 3/1 and consider the r=3 as the hypotenuse, 
which is ALWAYS taken positive, and y=1 as the opposite.  We use the Pythagorean theorem:

(hypotenuse)² = (adjacent)² + (opposite)²
           r² = x² + y²
           3² = x² + 1²
            9 = x² + 1
            8 = x²
          ±√8 = x
         ±2√2 = x

In QII the adjacent = x is negative, so we take the negative sign
x = -2√2.




adjacent = x = -2√2,  opposite = y = 1, hypotenuse = r = 3

sin(t) = opposite/hypotenuse = y/r = 1/3
cos(t) = adjacent/hypotenuse = x/r = -2√2/3 
tan(t) = opposite/adjacent = y/x = 1/(-2√2) = [1/(-2√2)][√2/√2] = 
       = (√2)/(-2∙2) = -√2/4 
sec(t) = hypotenuse/adjacent = r/x = 3/[-2√2] = [3/(-2√2)][√2/√2] = 
       = (3√2)/(-2∙2) = -3√2/4
csc(t) = hypotenuse/opposite = r/y = 3/1 = 3 (given!)
cot(t) = adjacent/opposite = x/y = -2√2/1 = -2√2

Edwin