I'm trying to show that if $\alpha$ is a regular curve parametrized by arc lenght whose range lies on the unit sphere centered at the origin, then $\kappa (s) = \sqrt{1+j^2}$ and $\tau (s) = \dfrac{j'(s)}{1+j^2(s)}$
where $j(s)=\det[\alpha (s), \alpha '(s), \alpha ''(s)]$. Any ideas?
Observe that since $\alpha$ is parametrized by arc length, we have $T=\alpha'$, and $\alpha'' = \kappa N$. Then $\alpha = aT+bN+cB$, for some $a,b,c$. However, note that $a=0$, since $\alpha$ has unit length, so the tangent vector is always orthogonal to $\alpha$. Thus $\alpha = bN+cB$, with $b^2+c^2=1$. Then we can differentiate and apply the Frenet-Serret formulas to get $$\alpha' = (-\kappa b)T +(b' -\tau c)N + (c'+\tau b)B = T,$$ so we get the equations $$1=-\kappa b,\quad b'=\tau c,\,\,\text{ and }\,\,c' = -\tau b.$$
Then taking the determinant $$\det (\alpha,\alpha', \alpha'') = \det(\alpha,T,\kappa N) = \kappa c \det(B,T,N) = \kappa c.$$
Observe that $1=1^2=\kappa^2b^2$, and $j^2 =\kappa^2 c^2$, so $$1+j^2 = \kappa^2(b^2+c^2)=\kappa^2,$$ so $\kappa = \sqrt{1+j^2}.$
As for $\tau$, note that $$c = \frac{j}{\sqrt{1+j^2}}.$$ Differentiating, we get $$c' = j'\frac{\sqrt{1+j^2}-j^2/\sqrt{1+j^2}}{1+j^2}= \frac{j'\kappa(1- c^2)}{1+j^2} = \frac{j'\kappa b^2}{1+j^2}.$$
Then recalling the equations we derived before, $\kappa b = -1$, $c'=-\tau b$, we obtain $$-\tau b = -\frac{j'b}{1+j^2},$$ so $$ \tau = \frac{j'}{1+j^2},$$ as desired.
