Answer
(a) $\frac{\pi}{6}$
(b) $P(-\frac{\sqrt 3}{2},\frac{1}{2})$
(c) $sin(t)=\frac{1}{2}, cos(t)=-\frac{\sqrt 3}{2}, tan(t)=-\frac{\sqrt 3}{3}, cot(t)=-\sqrt 3, sec(t)=-\frac{2\sqrt 3}{3},csc(t)=2$
Work Step by Step
(a) Start from $+x$ axis and go anti-clockwise to reach $t=-\frac{7\pi}{6}$, we can find that it will end up
in Quadrant II, and the reference number for t is $\bar t=\frac{\pi}{6}$
(b) Based on the results from (a) and table 4 on page 404 in the book, we have $P(-\frac{\sqrt 3}{2},\frac{1}{2})$
(c) Use the results from above, $sin(t)=y=\frac{1}{2}, cos(t)=x=-\frac{\sqrt 3}{2}, tan(t)=\frac{y}{x}=-\frac{\sqrt 3}{3}, cot(t)=\frac{x}{y}=-\sqrt 3, sec(t)=\frac{1}{x}=-\frac{2\sqrt 3}{3},csc(t)=\frac{1}{y}=2$
