Answer
(a) $\frac{\pi}{3}$
(b) $P(-\frac{1}{2},\frac{\sqrt 3}{2})$
(c) $sin(t)=\frac{\sqrt 3}{2},cos(t)=-\frac{1}{2},
tan(t)=-\sqrt 3,cot(t)=-\frac{\sqrt 3}{3},sec(t)=-2,csc(t)=\frac{2\sqrt 3}{3}$
Work Step by Step
(a) Since $+y$ axis is at $\frac{\pi}{2}$ and $-x$ axis is at $\pi$, $\frac{\pi}{2}\lt t\lt \pi$ indicates that
$t$ is in Quadrant II and $\bar t=\pi-t=\frac{\pi}{3}$
(b) The coordinates of point with $\bar t=\frac{\pi}{3}$ in Quadrant I is $(\frac{1}{2},\frac{\sqrt 3}{2})$, since point
$P$ is in Quadrant II, we have $P(-\frac{1}{2},\frac{\sqrt 3}{2})$
(c) With the above results, we have $sin(t)=y=\frac{\sqrt 3}{2},cos(t)=x=-\frac{1}{2},
tan(t)=\frac{y}{x}=-\sqrt 3,cot(t)=\frac{x}{y}=-\frac{\sqrt 3}{3},sec(t)=\frac{1}{x}=-2,csc(t)=\frac{1}{y}=\frac{2\sqrt 3}{3}$
