using
\(\sin^2 x + \cos^2 x = 1\)
\(\sin^2 \theta + (-3/5)^2 = 1\)
\(\sin^2 \theta + 9/25 = 1\)
\(\sin^2 \theta = 1-9/25 = (25-9)/25 = 16/25\)
\(\sin \theta = \pm 4/5\)
in 3rd quadrant, sin is negative.
so
\(\sin \theta = -4/5\)
now,
csc theta = 1/ sin theta = -5/4
tan theta = sin theta/ cos theta = -4/5 / -3/5 = + 4/3
