Answer
(a) $x = cot~t$
$y = csc~t$
We can see the graph below.
(b) $y = \sqrt{x^2+1}$
Work Step by Step
(a) $x = cot~t$
$y = csc~t$
for $t$ in $(0,\pi)$
When $t = \frac{\pi}{10}$:
$x = cot~\frac{\pi}{10} = 3.08$
$y = csc~\frac{\pi}{10} = 3.24$
When $t = \frac{\pi}{6}$:
$x = cot~\frac{\pi}{6} = 1.73$
$y = csc~\frac{\pi}{6} = 2$
When $t = \frac{\pi}{4}$:
$x = cot~\frac{\pi}{4} = 1$
$y = csc~\frac{\pi}{4} = \sqrt{2}$
When $t = \frac{\pi}{3}$:
$x = cot~\frac{\pi}{3} = 0.58$
$y = csc~\frac{\pi}{3} = 1.15$
When $t = \frac{\pi}{2}$:
$x = cot~\frac{\pi}{2} = 0$
$y = csc~\frac{\pi}{2} = 1$
When $t = \frac{2\pi}{3}$:
$x = cot~\frac{2\pi}{3} = -0.58$
$y = csc~\frac{2\pi}{3} = 1.15$
When $t = \frac{3\pi}{4}$:
$x = cot~\frac{3\pi}{4} = -1$
$y = csc~\frac{3\pi}{4} = \sqrt{2}$
When $t = \frac{5\pi}{6}$:
$x = cot~\frac{5\pi}{6} = -1.73$
$y = csc~\frac{5\pi}{6} = 2$
When $t = \frac{9\pi}{10}$:
$x = cot~\frac{9\pi}{10} = -3.08$
$y = csc~\frac{9\pi}{10} = 3.24$
We can see the graph below.
Note that $y=x$ is an asymptote for positive values of $x$ and $y=-x$ is an asymptote for negative values of $x$.
(b) $x = cot~t$
$x^2 = cot^2~t$
$y = csc~t$
$y^2 = csc^2~t$
$csc^2~t = cot^2~t + 1$
$y^2 = x^2 + 1$
$y = \sqrt{x^2+1}$
