$\gamma$ is line of curvature $\iff$ $\frac{d\theta}{ds} + \tau = 0$

Question

I have the following problem in differential geometry.

Suppose $S\in\mathbb R^3$ regular surface, and $\gamma:[0,L]\rightarrow S$ regular curve parametrized by arc-length.
Suppose that $\theta(s)$ is the angle between normal of the curve $n(s)$ and normal of the surface $N(\gamma(s))$.
I need to show that $\gamma$ is line of curvature $\iff$ $\frac{d\theta}{ds} + \tau = 0$ when $\tau$ is curve torsion.

my problem is that I don't know how to caculate $\frac{d\theta}{ds}$.

P.s.
I found here that this equality is called geodesic torsion

My attempt:
(Updated solution thanks to @TedShifrin)

I know that $cos\theta=<n,N>$ and by differentiating this equation I get:
$-sin\theta*\frac{d\theta}{ds} = <n\prime, N> + <n, N\prime>$.
Now using Fernet-serret I get:
$-sin\theta*\frac{d\theta}{ds} = <-\kappa\gamma\prime + \tau b(s), N> + <N\prime, n> = <\kappa\gamma\prime, N> + \tau(s)*sin\theta + <N\prime,n>$ (and this is because I know that the bi-normal is perpendicular to the curve normal).
Also I know that $Sp(\gamma\prime) = k_i \gamma\prime$ because it is line of curvature, and that $<\gamma\prime,N(\gamma)> = 0$ because $\gamma\prime \in TpS$.

Putting all together I get that:
$-sin\theta*\frac{d\theta}{ds} = \tau(s)*sin\theta$ and in case locally we have that $\theta \neq \pi*n$ we can divide by $sin\theta$ and get what we desired.

Answer 1

HINT: Differentiate $N\cdot n = \cos\theta$.