Answer
(a) See the graph.
(b) $y^2-x^2=1,~~x\gt0,~~y\gt0$
Work Step by Step
(a)
$t=\frac{\pi}{6}$:
$x=tan\frac{\pi}{6}=\frac{\sqrt 3}{3}$
$y=cot\frac{\pi}{6}=\sqrt 3$
Point: $(\frac{\sqrt 3}{3},\sqrt 3)$
$t=\frac{\pi}{4}$:
$x=tan\frac{\pi}{4}=1$
$y=cot\frac{\pi}{4}=1$
Point: $(1,1)$
$t=\frac{\pi}{3}$:
$x=tan\frac{\pi}{3}=\sqrt 3$
$y=cot\frac{\pi}{3}=\frac{\sqrt 3}{3}$
Point: $(\sqrt 3,\frac{\sqrt 3}{3})$
When $t→0$, $x→0$ and $y→∞$.
When $t→\frac{\pi}{2}$, $x→∞$ and $y→0$. So, $x\geq0$ and $y\geq0$
(b)
$x=tan~t$
$y=cot~t$
$y=\frac{1}{tan~t}$
$y=\frac{1}{x}$
$csc^2t=cot^2t+1$
$y^2=x^2+1$
$y^2-x^2=1$
