How can I find the implicit equation given $r=\langle\sec(u)\cos(v), \sec(u)\sin(v), \tan(u)\rangle$? 
Given $r=\langle\sec(u)\cos(v), \sec(u)\sin(v), \tan(u)\rangle$, how can I prove algebraically that it is a hyperboloid $x^2+y^2-z^2=1$?

I know that:
$$\begin{align*}
x &= \sec u \cos v \\
y &= \sec u \sin v \\
z &= \tan u
\end{align*}$$
But I can't wrap my mind around it.
EDIT:
I think I probably wasn't being clear: I meant given $r=\langle\sec(u)\cos(v), \sec(u)\sin(v), \tan(u)\rangle$, how can I find the implicit equation?
 A: With
$x = \sec(u)\cos(v), \tag{1}$
$y=\sec(u)\sin(v), \tag{2}$
and
$z=\tan(u), \tag{3}$,
we have
$x^2 + y^2 =  \sec^2(u)\cos^2(v) + \sec^2(u)\sin^2(v)$
$= \sec^2(u) (\cos^2(v) + \sin^2(v)) = (\sec^2(u))(1) = \sec^2(u); \tag{4}$
then
$x^2 + y^2 - z^2 = \sec^2(u) - \tan^2(u) = 1, \tag{5}$
using the identity $1 + \tan^2 (\theta) = \sec^2 (\theta)$.
That's pretty algebraical, ain't it?
If you want, as Dr. Shifrin mentions in his comment, to show the parametrization (1)-(3) yields the whole hyperboloid, then of course, as he also points out, find the unique $u$ such that $z = \tan(u)$; this is possible for all $z \in \Bbb R$ since $\tan(u)$ "goes from" $-\infty$ to $+\infty$ as $-\pi/2 \to u \to \pi/2$ and $(\tan(u))' = \sec^2(u) > 0$; then, having $u$, find that value of $v$ such that 
$(\cos (v), \sin(v)) = (\sqrt{1 + \tan^2(u)})^{-1}(x, y) = \vert \sec^{-1}(u) \vert (x, y) = \vert \cos(u) \vert (x, y) \tag{6}$
for $(x, y)$ such that $x^2 + y^2 = 1 + \tan^2(u) = 1 + z^2$.
Note Added Monday 29 December 2014 11:01 PM PST:  We can think of the equation $x^2 + y^2 - z^2 = 1$ as implicitly defining the hyperboloid as the level surface of the function $F(x, y, z) = x^2 + y^2 - z^2$ corresponding to the value $F(x, y, z) = 1$; since $\nabla F(x, y, z) = (2x, 2y, -2z)^T \ne 0$ anywhere on the set $\mathcal S = \{(x, y, z) \in \Bbb R^3 \mid F(x, y, z) = x^2 + y^2 - z^2 = 1 \}$ (indeed, we see that $\nabla F = 0$ only if $x = y = z = 0$, and $(0, 0, 0) \notin \mathcal S$), the implicit function theorem guarantees that the surface at all points may be expressed by an equation of the form, say, $z = g(x, y)$ (wherever $\partial F/\partial z \ne 0$; that is, wherever $z \ne 0$, with similar equations whenever $x$ or $y$ is nonvanishing).  Indeed, we may write $g(x, y)$ explicitly as $g(x, y) = \pm \sqrt{x^2 + y^2 - 1}$.  Similar formulas express $x$ as a function of $y$ and $z$ or $y$ in terms of $x$ and $z$ wherever the corresponding component of $\nabla F$ is nonzero.  Indeed, $z = \pm \sqrt{x^2 + y^2 - 1}$ is an explicit, parametric representation of the surface $\mathcal S$ with parameters $x$ and $y$, just as (1)-(3) form an explicit representation of $\mathcal S$ with parameters $u$ and $v$.  End of Note.
Another Note Added Monday 29 December 2014 11:53 PM PST:  In light of one of my comments on this topic, I should add that in the present case we can eliminate $u$ and $v$ from the parametric representation with relative ease:  noting that, by (3), $z^2 = \tan^2 (u)$, we have $\sec u = \sqrt{1 + \tan^2(u)} = \sqrt{1+ z^2}$; then from (1) and (2), $\cos (v) = x/\sqrt{1 + z^2}$, $\sin (v) = y/\sqrt{1 + z^2}$, and using $\cos^2(v) + \sin^2(v) = 1$ yields $x^2 + y^2 - z^2 = 1$ after a little bit of algebraic maneuvering.  This approach is but a slight twist on what I did in my original answer; the transformation works so well here by virtue of the specific forms of (1)-(3).  Would things were always so easy!  End of Note.
Hope this helps.  Cheers!
And as ever,
Fiat Lux!!!
