Answer
$y=x-1$ for $1\leq x\leq 2$
See graph
Work Step by Step
We are given the curve in parametric form:
$\begin{cases}
x=\sec^2 t\\
y=\tan^2 t
\end{cases}$
with $0\leq t\leq \dfrac{\pi}{4}$
Build a table of values:
$t=0; x=1; y=0; (x,y)=(1,0)$
$t=\dfrac{\pi}{16}; x\approx 1.04; y\approx 0.04; (x,y)=(1.04,0.04)$
$t=\dfrac{\pi}{8}; x\approx 1.71; y\approx 0.71; (x,y)=(1.71,0.71)$
$t=\dfrac{3\pi}{16}; x\approx 1.45; y\approx 0.45; (x,y)=(1.45,0.45)$
$t=\dfrac{\pi}{4}; x=2; y=1; (x,y)=(2,1)$
Plot the points and join them to graph the curve:
Determine the rectangular form of the equation:
$\tan^2 t+1=\sec^2 t$
$y+1=x$
$y=x-1$ for $1\leq x\leq 2$
