Prove that the length of the segment of the tangent of the tractrix between the point of tangency and the y axis is constantly equal to $1$. Let $\alpha: (0,π) \rightarrow \mathbb {R}^2$ be given by $\alpha(t) = (\sin t, \cos t + \log \tan\frac {t}{2})$,
where $t$ is the angle that the $y$ axis makes with the vector $\alpha'(t)$. The trace of $\alpha $ is called the tractrix. Show that
a. $\alpha$ is a differentiable parametrized curve, regular except at $t = π/2$.
b. The length of the segment of the tangent of the tractrix between the
point of tangency and the y axis is constantly equal to $1$.
My attempt:
My attempt for part a. is as follows:
I have that $\alpha'(t) = \left(\cos t, -\sin t + \frac {\frac {1}{2} \sec^2 \frac{t}{2}}{\tan \frac {t}{2}} \right)  = \left(\cos t, -\sin t + \frac {1}{2} \cot(\frac {t}{2}) \sec^2 (\frac {t}{2}) \right)$, So $\alpha $ is differentiable for all $t \in (0, \pi) $ because each of the components of $ \alpha $ have derivatives of all orders. On the other hand, a differentiable parametrized curve $ \beta: I \rightarrow \mathbb{R}^2 $ is said to be regular if $ \beta'(t) \neq 0 $ for all $ t \in I $. We have the particular case $ \cos t \neq 0 $ for all $ t \in (0, \pi) - \{\frac{\pi}{2}\} $, which implies that $ \alpha'(t) \neq 0 $ for all $ t \in (0, \pi) - \{\frac {\pi}{2} \} $. However $\alpha'(\frac {\pi}{2}) = (0,0)= \textbf{0 } $. Which shows that $\alpha$ is regular except at $ t = \frac{\pi}{2} $.  Is this reasoning correct?
For the part b. I need some suggestion because I have not been able to prove that part. I need some help to do that please.

 A: Part a: very good.
Part b: the tangent line at $\alpha(t)$ is parametrized by $\ell_t(s) = \alpha(t) + s\alpha'(t)$, so find the special $s_0$ for which the first component of $\ell_t(s_0)$ vanishes, and look at $f(t) = \langle \ell_t(s_0) - \alpha(t), \ell_t(s_0) - \alpha(t)\rangle$. With concrete formulas in hand, show that $f'(t) = 0$ and find $t_0$ such that $f(t_0) =1$.
A: Differentiate with respect to $t$ as parameter which is the angle the tangent makes to the y-axis as shown.  $\phi $ is angle made to horizontal x-axis. Is fully differentiable, including at cusp point $t=\pi/2$ where the curvature is infinite. For the unit speed the radial velocity vanishes at this cusp, similar to a situation when an automobile while reversing stops momentarily before continuing forward.
The continuity/smoothness  is more clear in  the 3 dimensional model as meridian of a pseudosphere.
I shall continue from your first line tangent vector where you  started correctly. Possible to calculate by trig from the given parametrization.

$$ \dfrac{dy}{dx}=\tan \phi=\dfrac{ \frac{d}{dt}\big(\cos t + \log \tan(t/2)\big)}{\frac{d}{dt}(\sin t)}$$
where $t$ is angle tangent makes to the vertical y-axis.
$$=\dfrac{-\sin t + \csc t}{\cos t}= \cot t $$
so that
$$ \phi= \pi/2-t$$
Angles $(\phi,t)$ are shown marked.
To find length between curve tangent point to y-axis $L$, the x-coordinate is given $\sin t $, and
$$L=\dfrac{ \sin t}{\cos \phi}=\dfrac{ \sin t}{\sin t}=1$$
There is a minor error in text drawing for $t$ at the top where I hope is shown here more clearly.
