For a unit-speed curve $\beta$ with unit tangent vector field $T_\beta$, we define the spherical image $\sigma$ With $\sigma(s)=T_\beta (s)$.
How can I prove that $$\kappa_\sigma=\sqrt{1+(\frac{\tau_\beta}{\kappa_\beta})^2}$$ and $$\tau_\sigma= \frac{(\frac{\tau_\beta}{\kappa_\beta})’}{\kappa_\beta(1+(\frac{\tau_\beta}{\kappa_\beta})^2)}$$
Here is what I have come up with so far:
$\sigma’(s)=T_\beta’(s)=\kappa_\beta N_\beta=T_\sigma(s)$ $$\implies T_\sigma’= \sigma’’(s)= \kappa_\beta’N_\beta +\kappa_\beta N_\beta’$$ $$=\kappa_\beta’N_\beta+\kappa_\beta(-\kappa_\beta T_\beta +\tau_\beta B_\beta)$$ $$=\kappa_\beta’ N_\beta +\kappa_\beta \tau_\beta B_\beta -\kappa_\beta^2 T_\beta$$
(Because $\tau_\beta=-B_\beta’\bullet N_\beta$) $$\implies \sigma’’(s)=\kappa_\beta’ N_\beta -\kappa_\beta^2 T_\beta$$
Then is: $\kappa_\sigma=\frac{\Vert \sigma’\times \sigma’’ \Vert}{v_\beta^3}=\Vert \sigma’\times \sigma’’ \Vert$ $$=\Vert (\kappa_\beta N_\beta) \times (\kappa_\beta’N_\beta -\kappa_\beta^2 T_\beta)\Vert$$ $$=\Vert(\kappa_\beta N_\beta) \times (-\kappa_\beta T_\beta)\Vert= \Vert \kappa_\beta^3 B_\beta \Vert$$
Can somebody help me find my mistake and how I can solve this? Thank you in advance.