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.