If unit tangent vector $\bar{t}$ and binormal vector $\bar{b}$ make angle $\theta, \phi$ with a constant unit vector $\hat{a}$. So prove that $$\frac{\sin\theta}{\sin\phi}\cdot\frac{d\theta}{d\phi}=-\frac{\kappa}{\tau}$$
I am not sure how to approach the problem. But applying the frenet-serret formulae $$\frac{dt}{ds}=-\frac{\kappa}{\tau}\frac{d}{ds} \\ \frac{dt}{d\theta}\frac{d\theta}{ds}=-\frac{\kappa}{\tau}\frac{db}{d\phi}\frac{d\phi}{ds}$$
Now if I can show $dt/d\theta= \sin\theta;db/d\phi= \sin\phi$ it would be proved. However I am unable to see how or why that would be the case.
Note that $$ \cos \theta = \frac{\langle \bar t, \hat a\rangle}{\|\bar t\| \|\hat a\|} = \langle \bar t, \hat a\rangle $$ and similarly $\cos \phi = \langle \bar b, \hat a\rangle$. Then $$ \begin{align*} \frac{\sin \theta\, d\theta}{\sin \phi\, d\phi} &= \frac{d(\cos \theta)}{d(\cos \phi)} \\ &= \frac{d( \langle \bar t, \hat a \rangle)}{d(\langle \bar b, \hat a \rangle)} \\ &= \frac{\langle d(\bar t), \hat a \rangle}{\langle d(\bar b), \hat a \rangle} \\ &= \frac{\langle \bar t', \hat a \rangle}{\langle \bar b', \hat a \rangle}. \end{align*} $$ To find the answer, apply the Frenet-Serret formulas for $\bar t$ and $\bar b$.