Help with trig identities needed

Given that sin x = -3/5, evaluate tan x/2. I know that I have to use trig identities and that
tan x/2 = (1 - cos x)/ (sin x), but I get lost and I don't understanding it completely. I don't get any credit for this question, but I need help understanding how to do it.

sin x = -3/5
evaluate tan(x/2)
▬▬▬▬▬▬▬▬▬▬▬ we know that tan (x/2) = (1 - cos x)/ (sin x)

sin x = -3/5
This tells us that the angle is in the III or IV quadrant.
Did the problem come with additional information?
If not, we must consider both quadrants.

▬▬▬▬▬▬▬▬▬▬▬ if angle x is in quadrant III
by way of the pythagorean theorem
sin x = -3/5
cos x = -4/5

tan (x/2) = (1 - cos x) / (sin x)
tan (x/2) = (1 - (-4/5) / (-3/5)
tan (x/2) = (9/5) / (-3/5)
tan (x/2) = -3

▬▬▬▬▬▬▬▬▬▬▬ if angle x is in quadrant IV
by way of the pythagorean theorem
sin x = -3/5
cos x = 4/5

tan (x/2) = (1 - cos x) / (sin x)
tan (x/2) = (1 - (4/5) / (-3/5)
tan (x/2) = (1/5) / (-3/5)
tan (x/2) = -⅓

I hope this helps you understand.