Hello, kevin11!
(a) Eliminate parameter $\displaystyle t$ to find the equation in rectangular form.
(b) Sketch the curve defined by: .$\displaystyle \begin{Bmatrix}x &=& \sec t \\ y &=& \tan^2\!t\end{Bmatrix}\quad 0\leq t \leq \pi $ skeeter is absolutely correct!
The equation is: .$\displaystyle y \:=\:x^21\quad\hdots$ a parabola.
. . It opens upward, vertex (0,1), xintercepts (±1, 0)
But there's more . . .
I too made a chart and got these values (some approximate).
. . $\displaystyle \begin{array}{ccc}
t & x & y \\ \hline \\[4mm]
0 & 1 & 0 \\ \\[4mm]
\frac{\pi}{6} & 1.15 & 0.33 \\ \\[4mm]
\frac{\pi}{4} & 1.41 & 1 \\ \\[4mm]
\frac{\pi}{3} & 2 & 3 \end{array}$
L . $\displaystyle \begin{array}{ccc}
\frac{\pi}{2} & \infty & \infty \\ \\[4mm]
\frac{2\pi}{3} & \text{}2 & 3 \\ \\[4mm]
\frac{3\pi}{4} & \text{}1.41 & 1 \\ \\[4mm]
\frac{5\pi}{6} & \text{}1.15 & 0.33 \\ \\[4mm]
\pi & \text{}1 & 0 \end{array}$
So the graph looks like this:
Code:

* 3+ *

*  *
*  *
    *   +   *    
2 1  1 2
