Let $\alpha$ be a unit speed curve. Its tangent indicatrix $\sigma$ is defined by $\sigma(t)=T(t)$. Find torsion and curvature of $\sigma$ with respect to the torsion and curvature of $\alpha$.
There is an answer on my book. But I don't understand (*) part
$$ \kappa_\sigma=\frac{\|\sigma'\times \sigma''\|}{\|\sigma'\|^3}$$
$$ \tau_\sigma=\frac{\left<\sigma'\times\sigma'',\sigma'''\right>}{\|\alpha'\times\alpha''\|^2} $$
$$\sigma'(t)=T'(t) \Rightarrow \sigma'=T'=\kappa N$$
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We find $\sigma'\times\sigma''=\kappa^3B+\kappa^2\tau T$
$$\|\sigma'\|=\sqrt{\left<\sigma',\sigma'\right>}=\sqrt{\left<\kappa N,\kappa N\right>}={\kappa^2\left<N,N\right>}=|\kappa|=\|\alpha''\|$$
$$\|\sigma'\times\sigma''\|=\sqrt{(\kappa^2\tau)^2+(\kappa^3)^2}=\kappa^2\sqrt{\tau^2+\kappa^2} \tag{*}$$
$$\kappa_\sigma=\frac{\sqrt{\tau^2+\kappa^2}}{\kappa}$$
...
I couldn't understand the equation (*). Can you explain it please? (The expression "tangents indicator" may not be well translated. I translated into English and $\kappa$ is the curvature and $\tau$ , $T$ tangent vector, and $B=T\times N$.)