Answer
(a) See the graph.
(b) $y^2-x^2=1,~~y\geq1$
Work Step by Step
(a)
$t=\frac{\pi}{4}$:
$x=cot\frac{\pi}{4}=1$
$y=csc\frac{\pi}{4}=\sqrt 2$
Point: $(1,\sqrt 2)$
$t=\frac{\pi}{2}$:
$x=cot\frac{\pi}{2}=0$
$y=csc\frac{\pi}{4}=1$
Point: $(0,1)$
$t=\frac{3\pi}{4}$:
$x=cot\frac{3\pi}{4}=-1$
$y=csc\frac{3\pi}{4}=\sqrt 2$
Point: $(-1,\sqrt 2)$
When $t→0$ both $x$ and $y$ $→∞$. So, $x\geq1$ and $y\geq0$
When $t→\pi$ both $x$ and $y$ $→-∞$. So, $x\geq1$ and $y\geq0$
(b)
$x=cot~t$
$x^2=cot^2t$
$y=csc~t$
$y^2=csc^2t$
$csc^2t=cot^2t+1$
$y^2=x^2+1$
$y^2-x^2=1$
