What is the space curve with curvature and torsion obeying $ \kappa = \cos s, \tau  = \sin s  $ and  passing through (1,0,0), TNB triad identity matrix?
previous link
When numerically computed it looks like a catenoid surface of revolution for all appearances. ( unit curvature at extremes and like an symptote at symmetrical plane). Its parametrization is not mentioned in Differential Geometry books that I could come across. 
EDIT 1:


*

*Can it at least be proved that the line lies on a surface of revolution?

*In general for a surface of revolution, what functional relation/s can exist between $ \kappa, \tau $ ?
EDIT 2:
Initial values for (T,N,B) taken as an identity matrix to fix rotation.
EDIT 3: 
The lines given by  achille hui are described on an equiangular rotational hyperboloid inclined at $\pi/4$  to one axis, result of initial point $(0,0,0)$ and starting $T,N,B$ an identity matrix.
May I request someone to extend it further or integrate the locus incorporating constants for generality ?  like
$$ (a \kappa, a \tau)  = ( \cos s/b , \sin s/b)  $$ 
An image of a hyperboloid of revolution for $ (a= 1/6, b =2 ) $:

Motivation of this post was in fact to readily plot these lines on a surface of revolution using scalar curvatures properly.
EDIT 4:
In hindsight it slightly puzzles (me)... Although we started off with two intrinsic curvature invariants in 3-space, when combined it forces the arc to belong to 1- sheet hyperboloid  $ of \, revolution $ in $ \mathbb R^2 $ that was neither specified in the input nor expected (by me). Is there some explanation or insight for it to so happen?
 A: Instead of a curve passing through $(1,0,0)$, choose the coordinate system such that at time $t = 0$, the curve start at $(0,0,0)$ with $\vec{T}, \vec{N}, \vec{B}$ in the direction of $x, y, z$ respectively.
It turns out for $\kappa(s) = \cos(s), \tau(s) = \sin(s)$, one can convert
the ODE for Fernet-Serret frame to an ODE over quaternion and solve it explicitly.
The derivation is pretty messy and I won't reproduce it here. Feel free to throw
the expression of space curve below to a CAS and verify it works.
In any event, the space curve $\vec{X}(s) = (x(s),y(s),z(s))$ is given by the formula:
$$\begin{cases}
x(s) &= \frac{3}{\sqrt{2}}\sin(\sqrt{2}s)\cos(s) - 2\cos(\sqrt{2}s)\sin(s)\\
y(s) &= -\frac12(3\cos(\sqrt{2}s) + 1)\cos(s) - \sqrt{2}\sin(\sqrt{2}s)\sin(s) + 2\\
z(s) &= +\frac12(3\cos(\sqrt{2}s)-1)\cos(s) + \sqrt{2}\sin(\sqrt{2}s)\sin(s) - 1
\end{cases}
$$
With a little bit of algebra (or an CAS), one can verify this curve lies on a one-sheet hyperboloid:
$$x^2 - 2(y-2)(z+1) = 4$$
Update
For the general case $\kappa(s) = \rho\cos(\alpha s), \tau(s) = \rho\sin(\alpha s)$ where $\rho, \alpha$ are constants, 
if I didn't make any mistake, the space curve is given by
$$\begin{cases}
x(s) &= \frac{\rho^2}{2\beta} S(s)\\
y(s) &= y_0 -\frac{\rho^2}{2\beta^2}\left( \frac{2}{\rho}\cos(\alpha s) + \rho C(s)\right)\\
z(s) &= z_0 -\frac{\rho^2}{2\beta^2}\left( \frac{2}{\alpha}\cos(\alpha s) - \alpha C(s)\right) 
\end{cases}
$$
where $\beta = \sqrt{\alpha^2+\rho^2}$, 
$y_0 = \frac{2}{\rho}$, 
$z_0 = \frac{\rho^2-2\alpha^2}{a\rho^2}$ and
$\displaystyle\;
\begin{cases}
C(s) &= 
\frac{\cos((\beta-\alpha)s)}{(\beta-\alpha)^2} +
\frac{\cos((\beta+\alpha)s)}{(\beta+\alpha)^2}\\
S(s) &= 
\frac{\sin((\beta-\alpha)s)}{(\beta-\alpha)^2} +
\frac{\sin((\beta+\alpha)s)}{(\beta+\alpha)^2}\\
\end{cases}
$.
Furthermore, the space curve lies on the one-sheet hyperboloid:
$$\rho^2\beta^2 x^2 + \rho^2(\alpha(z-z_0) - \rho(y-y_0))^2 - \alpha^2(\rho(z-z_0) + \alpha(y-y_0))^2 = \frac{4\alpha^2\beta^2}{\rho^2}$$
A: Since curvature and torsion are invariant under Euclidean transformations, you can take any curve with given $\kappa(s), \tau(s)$ and revolve around any line, obtaining a surface of revolution made up of curves with the same $\kappa(s)$ and $\tau(s)$.  But given an initial point on the curve where, to keep things simple, $\kappa$ and $\tau$ are nonzero, the initial  $(T,N,B)$ could point in any three orthogonal directions, so the curves don't all lie on a surface, they fill up a region of space. 
