Torsion and curvature of generalized helix A regular curve $\textbf{$\gamma$}$ in $\mathbb{R}^3$ with curvature $> 0$ is called a generalized helix if its tangent vector makes a fixed angle $\theta$ with a fixed unit vector $\textbf{a}$. Show that the torsion $\tau$ and curvature $\kappa$ of $\textbf{$\gamma$}$ are related by $\tau = ±\kappa \cot \theta$. Show conversely that, if the torsion and curvature of a regular curve are related by $\tau = \lambda \kappa$ where $\lambda$ is a constant, then the curve is a generalized helix. Note that we have the Frenet equations $\textbf{t}'=\kappa \textbf{n}$ and $\textbf{b}'= -\tau \textbf{n}$.
I have done the first half of the proof. In the second half, I claim that $\textbf{a}=\textbf{t} \cos \theta ± \textbf{b}\sin \theta$ where $\textbf{t} $ and $\textbf{b}$ are the unit tangent and binormal vectors, respectively, satisfies the conditions for a general helix. I'm trying to show that the derivative of $\textbf{a}$ is $0$ if we assume that $\tau = \lambda \kappa$, proving that $\textbf{a}$ is constant. I tried showing that $\mathbf{a' \cdot a'}=0$, which would prove that $\mathbf{a'}=0$, but I was unable to make it work. Any suggestions or different ideas on how to approach the converse would be appreciated.
My attempt: 
$\textbf{a}'=\textbf{t}' \cos \theta ± \textbf{b}' \sin \theta 
=\kappa \textbf{n} \cos \theta ± \tau \textbf{n} \sin \theta = \kappa \textbf{n} \cos \theta ± \lambda \kappa \textbf{n} \sin \theta$
 A: Since $\mathbf a$ and $T$ are unit vectors we may write
$\mathbf a \cdot T = \Vert a \Vert \Vert T \Vert \cos \theta = \cos \theta, \tag 1$
where $\theta$ angle 'twixt $\mathbf a$ and $T$; we may differentiate this equation with respect to the arc-length $s$ along our curve $\gamma(s)$, yielding
$\dot {\mathbf a} \cdot T + \mathbf a \cdot \dot T = 0, \tag 2$
and since
$\dot {\mathbf a} = 0 \tag{2.5}$
and we have the first Frenet-Serret relation
$\dot T = \kappa N, \tag{2.6}$
(2) becomes
$\mathbf a \cdot \dot { \kappa N} = 0,  \tag 3$
and with
$\kappa > 0 \tag{3.5}$
we find that
$\mathbf a \cdot N = 0, \tag 4$
which we may differentiate yet again with respect to $s$:
$\dot{\mathbf a} \cdot N + \mathbf a \cdot {\dot N} = 0, \tag 5$
and yet again via (2.5) we may write
$\mathbf a \cdot {\dot N}  = 0; \tag{5.5}$
we now deploy the second Frenet-Serret equation
$\dot N = -\kappa T + \tau B \tag 6$
to obtain
$\mathbf a \cdot {(-\kappa T + \tau B)} = 0, \tag 7$
or
$-\kappa \mathbf a \cdot T + \tau \mathbf a \cdot B = 0, \tag 8$
whence, using (1),
$-\kappa \cos \theta + \tau \mathbf a  \cdot B = 0; \tag 9$
we expand $\mathbf a$ in terms of $T$, $N$, $B$ using (1) and (4) as follows:
$\mathbf a = (\mathbf a \cdot T)T +  (\mathbf a \cdot N)N +   (\mathbf a \cdot B)B = (\cos \theta) T +(\mathbf a \cdot B)B; \tag{10}$
since
$\Vert \mathbf a \Vert = \Vert T \Vert = \Vert B \Vert = 1, \tag{11}$
and
$T \cdot B = \mathbf a \cdot N = 0, \tag{12}$
we infer from (10) that
$1 = \Vert \mathbf a  \Vert^2 = \cos^2 \theta \Vert T \Vert^2 + (\mathbf a \cdot B)^2 \Vert B \Vert^2$
$= \cos^2 \theta + (\mathbf a \cdot B)^2, \tag{13}$
which implies that
$\mathbf a \cdot B = \pm \sin \theta; \tag{14}$
substituting this into (9) yields
$-\kappa \cos \theta \pm \tau \sin \theta = 0, \tag{15}$
and then
$\kappa \cos \theta = \pm \tau \sin \theta, \tag{16}$
whence
$\tau = \pm \kappa \cot \theta, \tag{17}$
as per request.
Going the other way, given that
$\tau = \lambda \kappa \tag{18}$
for some constant
$\lambda \in \Bbb R, \tag{19}$
we may choose $\theta$ such that
$\lambda = \cot \theta = \dfrac{\cos \theta}{\sin \theta}; \tag{20}$
next, we set
$\mathbf a = (\cos \theta) T + (\sin \theta) B, \tag{21}$
and note this implies;
$\Vert a \Vert = \sqrt{\cos^2 \theta \Vert T \Vert^2 + \sin^2 \theta \Vert B \Vert^2}$
$= \sqrt{\cos^2 \theta + \sin^2 \theta} = \sqrt 1 = 1; \tag{21.1}$
and apply  $d/ds$ to (21):
$\dot {\mathbf a} = (\cos \theta) \dot T + (\sin \theta) \dot B; \tag{22}$
we substitute (2.6) and the third Frenet-Serret equation
$\dot B = -\tau N, \tag{23}$
and obtain
$\dot{\mathbf a} = (\cos \theta)\kappa N - (\sin \theta)\tau N = (\kappa \cos \theta - \tau \sin \theta) N; \tag{24}$
in light  of (18) and (20),
$\tau = \lambda \kappa = \dfrac{\cos \theta}{\sin \theta} \kappa, \tag{25}$
and thus
$\tau \sin \theta = \kappa \cos \theta, \tag{26}$
which in concert with (24) shows that
$\dot{\mathbf a} = 0. \tag{27}$
We have shown the existence of a constant vector $\mathbf a$ and a constant angle $\theta$ such that (1) binds; $\gamma(s)$ is a generalized helix.
Note Added in Edit, Monday 20 January 2020 6:31 PM PST:  As we transit 'twixt (15) and (17), we have occasion to divide through by $\sin \theta$; thus we should address the question of just when
$\sin \theta = 0. \tag{28}$
Now (28) occurs precisely when
$\theta = 0, \pi, \tag{29}$
that is, when $T$ is aligned parallel or anti-parallel to $\mathbf a$. (We  observe that
$0 \le \theta \le \pi \tag{30}$
since it is the angle between the vectors $\mathbf a$ and $T$.)  But (29) implies
$T = \pm \mathbf a, \tag{31}$
which further implies that $\gamma(s)$ is a straight line; as such,
$\kappa = 0, \tag{32}$
which contradicts our hypothesis that the curvature of $\gamma(s)$ is positive.  Therefore
$\sin \theta \ne 0, \tag{33}$
and the quotient
$\cot \theta = \dfrac{\cos \theta}{\sin \theta} \tag{34}$
may legitimately be formed.  End of Note.
A: OK, now that you added the relevant information. You never specified how you would choose $\theta$. I.e., you haven't yet related $\lambda$ and $\theta$. Everything you typed suggested that you should take $\theta$ so that $\lambda = \cot\theta$. If you do that, what is $\cos\theta - \lambda\sin\theta$? Now simplify. (Obviously, you need the correct sign choice.)
