Question

Let theta be an angle in quadrant III such that cos theta=-3/5 .

Answer 1

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