Tractrix tangent segment (from Baby Do-Carmo) I need some help with question 4 in section 1.3 in Baby Do-Carmo textbook in DG. The question asks: Let $\alpha(t):(0,\pi)\rightarrow R^2 $ be given by: $$ \alpha(t)= (\cos(t), \cos(t) +\log(\tan(t/2)) $$ its image is called the tractrix. Question b, asks to prove that the length of the segment of the tangent of the tractrix between the point of tangency and the y axis is constantly 1. Now the angle between $\alpha$ and the y axis is t. So basically if I were to use the sine theorem from trig, where $$\frac{S}{\sin(t)} = \frac{|\alpha(t)|}{\sin(\pi-(t+\angle \alpha(t) \alpha '(t)))}$$ Where S is the required line segment I am looking for. Now I am only left with calculating the angle between $\alpha(t)$ and $\alpha '(t)$, is this about right, or I am way off here? It's hell of a calculation if I am right (and it's really rare when I am). Thanks.

You can find the tangent line of a curve at a point $\alpha(t)=(\alpha_1(t),\alpha_2(t))$ by the formula $$ \det\begin{pmatrix} X-\alpha_1(t) & \alpha_1'(t) \\\ Y-\alpha_2(t) & \alpha_2'(t)\end{pmatrix} $$ (for the sake of completeness, with your curve it is the locus of $(X,Y)$ such that $-\sin t \; (\log \tan \frac{t}{2} +X-Y) +X \csc t-\cot t$).

Now it's only a matter of computation of the length of the segment between $\alpha(t)$ and the $Y$-axis intercept of the former line. Am I right?