Answer
(a) see proof below.
(b) $-\frac{4}{5}, \frac{3}{5},-\frac{4}{3}$
Work Step by Step
(a) To show that point P is on the circle, plug in the coordinates into the equation $x^2+y^2=1$
we have $(\frac{3}{5})^2+(-\frac{4}{5})^2=\frac{9}{25}+\frac{16}{25}=1$, thus point P is on the unit circle.
(b) Base on the signs of the coordinates in point P, we see that point P is in Quadrant IV,
we have $sin(t)=y=-\frac{4}{5}, cos(t)=x=\frac{3}{5}, tan(t)=\frac{y}{x}=-\frac{4}{3}$
