If the unit tangent vector $\vec{t}$ and binonormal $\vec{b}$ makes an angle $\theta$ and $\phi$ respectively with a constant unit vector $\vec{a}$, prove that , $ \frac{sin\theta}{sin\phi}. \frac{d\theta}{d\phi}=\frac{\kappa}{\tau} \ $.
I have tried it by considering $\vec{t}$ , $\vec{b}$, $\vec{n}$ as orthonormal vectors. Since $\vec{t}$.$\vec{u}$ = $cos\theta$ and hence $\vec{n}$.$\vec{u}$ = $0$ and $\vec{b}$.$\vec{u}$ = $cos\phi$ and then i have put the value in $\vec{u}$ = <$\vec{t}$.$\vec{u}$> $\vec{t}$ + <$\vec{b}$.$\vec{u}$> $\vec{b}$ + <$\vec{n}$.$\vec{u}$> $\vec{n}$ then i have try to differentiate it but answer is not coming. How can i do after this ?
According to the sign convention I use, your equation is off by a factor of $-1$. You're making things too complicated. Assuming $\vec n\cdot\vec a\ne 0$, differentiate $\vec t\cdot\vec a = \cos\theta$ and $\vec b\cdot\vec a = \cos\phi$.
If $\vec n\cdot\vec a = 0$ identically, then $\theta$ and $\phi$ will both be constant and the curve will be a generalized helix, with $\tau/\kappa = \pm\cot\theta$ (and so the problem really does need to stipulate that $\vec n\cdot\vec a\ne 0$).