Answer
$y = \frac{3}{5}~\sqrt{x^2+25}$, where $x$ in $(-\infty, \infty)$
Work Step by Step
$x = 5~tan~t$
$x^2 = 25~tan^2~t$
$x^2 = 25~(sec^2~t-1)$
$sec^2~t = \frac{x^2}{25}+1$
$sec~t = \sqrt{\frac{x^2+25}{25}}$
We can replace this expression in the equation with $y$:
$y = 3~sec~t$
$y = 3~\sqrt{\frac{x^2+25}{25}}$
$y = \frac{3}{5}~\sqrt{x^2+25}$
Since $t$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$, then $x$ in $(-\infty, \infty)$
$y = \frac{3}{5}~\sqrt{x^2+25}$, where $x$ in $(-\infty, \infty)$
