Tractrix Tangent Length Problem I am having trouble with a problem I am working on

The trace of $\vec{r}(t):=\sin(t)\vec{i}+[\cos(t)+\ln[\tan(t/2)]]\vec{j}$ where $t\in(0,\pi)$ is called a tractrix. Show the length of the line segment of the tangent between the point of tangency on the tractrix and the y-axis is constantly equal to 1.

I've got a good visualization of the problem I am attempting. I need to find the equation for the tangent line by first finding $dy/dx$, with $x=f(t)=\sin(t)$ and $y=g(t)=\cos(t)+\ln[\tan(t/2)]$. From there I need to show that the length of the line is always equal to one. I know the formula for the length of a curve of a parametric function, but I don't know what to use as the beginning and ending points to show some general case.
 A: You can check the length of the tangent of a function at a given point is:
$T = \left|\dfrac{y}{y'}\sqrt{y'^2+1}\right|$
For full detail check Piskunov's Calculus, Vol I, page 127.
Since in your equation the tractrix is running along the $y$ axis I'll use it running through the $x$ axis (since the formula I give you is for such cases).
So you have
$y = \sin t$
$x = \cos t +\log \tan\dfrac{t}{2}$
$\dfrac{dy}{dt} = \cos t$ 
$\dfrac{dx}{dt} = -\sin t +\dfrac{\sec ^2\dfrac{t}{2}}{2\tan\dfrac{t}{2}}$
Using the double angle formula you'll get
$$\frac{{dx}}{{dt}} = \frac{1}{{\sin t}} - \sin t$$
or
$$\frac{{dx}}{{dt}} = \frac{{\cos^2 t}}{{\sin t}} $$
and then
$$\frac{{dy}}{{dx}} = \frac{{\cos t}}{{\dfrac{{{{\cos }^2}t}}{{\sin t}}}} = \tan t$$
Plugging this in gives
$$T = \left| {\dfrac{{\sin t}}{{\dfrac{{\sin t}}{{\cos t}}}}\sqrt {{{\tan }^2}t + 1} } \right| = \left| {\dfrac{{\sin t}}{{\sin t}}\cos t\sec t} \right| = 1$$

Finding the formulas.

From geometry we know that
$$\tan \theta = \dfrac{y}{S_T}$$
But since $\tan \theta = \dfrac{dy}{dx} = y'$ we can put
$$y' = \dfrac{y}{S_T}$$
$$S_T = \left|\dfrac{y}{y'}\right|$$
We consider the absolut value since $\tan \theta$ isn't always positive.
But now, knowing $S_T$ we can use
$$T^2 = S_T^2 +y^2$$
$$T^2 = \dfrac{y^2}{y'^2} +y^2$$
$$T = \left|y' \sqrt{\dfrac{1}{y'^2}+1}\right|$$
$$T = \left|\dfrac{y}{y'} \sqrt{{y'^2}+1}\right|$$
Looking to the other triangle, we get
$$\tan \theta = \dfrac{S_N}{y}$$
So
$$|y y'|= S_N$$
And finally since
$$N^2 = y^2 + S_N^2$$
You get
$$N^2 = y^2 + y^2 y'^2$$
$$N = \left| y\sqrt{1+y'^2}\right|$$
These lengths are called tangent, subtangent, normal and subnormal. Remember $y = f(x)$ and $y' = f'(x)$, so you always have to plug in a number to get a value.
