Deriving a relationship between curvature, torsion and the curvature of tangent vector Let $\alpha(s)$ be a regular and biregular curve in $\mathbb{R}^3$ parametrized in arc length with its curvature $k(s)$ and torsion $\tau(s)$. Then we can think of it's tangent vector $T(s)$ as another curve which range is contained on the unit sphere, let's call it $\beta(s)=\alpha'(s)$ with its curvature, say $k_1(s)$. I have to prove the following relation:
$$k_1^2=\frac {k^2+\tau^2}{k^2}$$
I've tried different approaches and I've derived some relations (maybe true as long as my calculation were right) using Frenet-Serres formula but I still cannot conclude.
 A: Since the following discussion addresses multiple curves $\gamma$ each with its own tangent, normal, and binormal vectors, I will distinguish 'twixt the Frenet-Serret apparatus of different curves via subscripting:  thus the unit tangent vector to $\gamma$ will be denoted by $T_\gamma$ and so forth.
We are considering the curve $\beta(s)$ given as the tangent vector $T_\alpha(s)$ to the unit speed curve $\alpha(s)$, so that
$\beta(s) = T_\alpha(s) = \alpha'(s) \tag 1$
with
$\langle \beta(s), \beta(s) \rangle = \langle T_\alpha(s), T_\alpha(s) \rangle = 1, \tag 2$
and we wish to compute the curvature $\kappa_1(s)$ of $\beta(s)$.  One way of addressing this is to express the Frenet-Serret apparatus of $\beta(s)$ in terms of $\alpha(s)$ and its Frenet frame $T_\alpha$, $N_\alpha$, $B_\alpha$.  We recall that $s$ is the arc-length along $\alpha(s)$; however, $s$ is not in general the arc-length along $\beta(s) = T_\alpha(s)$; that is, $\beta(s)$ is not in general a unit-speed curve.  We thus donote the arc-length along $\beta$ by $\sigma$; then $T_\beta$ is given by
$T_\beta(\sigma) = \beta'(\sigma) = \dfrac{dT_\alpha(s)}{ds}\dfrac{ds}{d\sigma} = \kappa(s)N_\alpha(s)\dfrac{ds}{d\sigma}, \tag 3$
and since
$\Vert T_\beta(\sigma) \Vert = \Vert N_\alpha(s) \Vert = 1, \tag 4$
it follows from (3) that
$\left \vert \kappa(s) \dfrac{ds}{d\sigma} \right \vert = \left \vert \kappa(s) \dfrac{ds}{d\sigma} \right \vert \Vert N_\alpha(s) \Vert = \left \Vert \kappa(s) \dfrac{ds}{d\sigma} N_\alpha(s) \right \Vert = \Vert T_\beta(\sigma) \Vert = 1;\tag 5$
we conclude from (5) that
$\kappa^2(s) \left (  \dfrac{ds}{d\sigma} \right )^2 = \left \vert \kappa(s) \dfrac{ds}{d\sigma} \right \vert^2 = 1, \tag 6$
a formula which we shall shortly deploy.  Since $\kappa(s) > 0$, we may also write
$\kappa(s) \left \vert \dfrac{ds}{d\sigma} \right \vert = \left \vert \kappa(s) \dfrac{ds}{d\sigma} \right \vert = 1, \tag 7$
and from this we see that
$\kappa(s)  \dfrac{ds}{d\sigma} = \pm 1; \tag 8$
by reversing the direction of traverse of $\beta(\sigma)$ with respect to $s$ if necessary we may also assume that $ds / d\sigma > 0$, so that in fact
$\kappa(s)  \dfrac{ds}{d\sigma} = 1, \tag 9$
which, though not further necessary here, is worth noting in it's own right.
We compute the curvature $\kappa_1(\sigma)$ of $\beta(\sigma)$ via the Frenet-Serret equations; as applied to $\beta(\sigma)$, we find via (3) that
$\kappa_1(\sigma) N_\beta(\sigma) = T_\beta'(\sigma) = \dfrac{d}{d\sigma} \left ( \kappa(s)N_\alpha(s)\dfrac{ds}{d\sigma} \right ); \tag{10}$
now,
$\dfrac{d}{d\sigma} \left ( \kappa(s)N_\alpha(s)\dfrac{ds}{d\sigma} \right ) = \dfrac{d}{d\sigma} \left ( \left (\kappa(s)\dfrac{ds}{d\sigma}\right )N_\alpha(s)\right )$
$= \dfrac{d}{d\sigma} \left (\kappa(s)\dfrac{ds}{d\sigma}\right ) N_\alpha(s) + \left ( \kappa(s)\dfrac{ds}{d\sigma}\right )\dfrac{d}{d\sigma} N_\alpha(s); \tag{11}$
from (8), we see that $(\kappa(s)ds / d\sigma)$ must be constant, whence
$\dfrac{d}{d\sigma} \left (\kappa(s)\dfrac{ds}{d\sigma}\right ) = 0; \tag{12}$
combining (8), (10), (11) and (12):
$\kappa_1(\sigma) N_\beta(\sigma) =\pm \dfrac{d}{d\sigma} N_\alpha(s); \tag{13}$
we proceed:
$\dfrac{dN_\alpha(s)}{d\sigma}  = \dfrac{dN_\alpha(s)}{ds} \dfrac{ds}{d\sigma} =  \dfrac{ds}{d\sigma} (-\kappa(s) T_\alpha(s) + \tau(s) B_\alpha(s)), \tag{14}$
by virtue of the Frenet-Serret equations for $\alpha(s)$; from (13) and(14),
$\kappa_1^2(\sigma) = (\kappa_1(\sigma) N_\beta(\sigma)) \cdot (\kappa_1(\sigma) N_\beta(\sigma))  = \left (\dfrac{d}{d\sigma} N_\alpha(s) \right) \cdot \left (\dfrac{d}{d\sigma} N_\alpha(s) \right)$
$= \left (\dfrac{ds}{d\sigma} \right )^2  (-\kappa(s) T_\alpha(s) + \tau(s) B_\alpha(s)) \cdot  (-\kappa(s) T_\alpha(s) + \tau(s) B_\alpha(s))$
$=  \left (\dfrac{ds}{d\sigma} \right )^2 (\kappa^2(s) + \tau^2(s)), \tag{15}$
where we have used $\Vert B_\alpha(s) \Vert = 1$, $T_\alpha(s) \cdot B_\alpha(s) = 0$ as well as (2) in this derivation.  Finally, via (6),
$\kappa_1^2(\sigma) = \dfrac{\kappa^2(s) + \tau^2(s)}{\kappa^2(s)}, \tag{16}$
the desired result.
