Answer
$-\frac{2}{15}$
Work Step by Step
Since the terminal point determined by t is in Quadrant III, we know that $sin(t)\lt0$.
Use Pythagorean Identify with $cos(t)=-\frac{8}{17}$, we have $sin(t)=-\sqrt {1-cos^2t}=-\frac{15}{17}$
Thus, $tan(t)cot(t)+csc(t)=\frac{sin(t)}{cos(t)}\times\frac{cos(t)}{sin(t)}+\frac{1}{sin(t)}=1-\frac{17}{15}=-\frac{2}{15}$
