The textbook I'm using define the tractrix by $ T=\{(\sin t, \cos t+\log (\tan (t/2))):0<t\leq\pi\}$ and define the pseudospher being the tractrix roting around the $z$-axis, I have to prove that the pseudosphere without a circumference is orientable.
I divided this in several steps:
$(1)$ I prove that $T= f\cup(-f)$ being $f(x)=\log\left( \frac{1+\sqrt{1-x^2}}{x}\right) - \sqrt{1-x^2}$
Let's say that $(x,y)\in T$, then $x=\sin t$ and $y=\cos t + \log (\tan (t/2))$. By the first equality is $y=\arcsin x$ then $y=\cos (\arcsin x)+ \log\left(\tan \left(\frac{1}{2}\arcsin x \right)\right)$.
Here is my I first issue, I can't prove that $$\tan \left(\frac{1}{2}\arcsin x \right) = \left( \frac{1+\sqrt{1-x^2}}{x}\right).$$ To do this I tried to write it as $\tan\left( \frac{\sin \left(\frac{\arcsin x}{2}\right)}{\cos\left(\frac{\arcsin x}{2} \right)} \right)$, then using the identities $$\sin \left(\frac{1}{2}\arcsin x\right) = \frac{\sqrt{x+1}}{2} - \frac{\sqrt{1-x}}{2}$$ $$\cos \left(\frac{1}{2}\arcsin x\right) = \frac{\sqrt{x+1}}{2} + \frac{\sqrt{1-x}}{2}$$
then I have $$\tan \left(\frac{1}{2}\arcsin x \right) = \frac{\sqrt{x+1}-\sqrt{1-x}}{\sqrt{x+1}+\sqrt{1-x}} \\ = \frac{(\sqrt{x+1}-\sqrt{1-x})(\sqrt{x+1}+\sqrt{1-x})}{(\sqrt{x+1}+\sqrt{1-x})^2}\\ = \frac{2x}{2+2\sqrt{1-x^2}}=\frac{x}{1+1\sqrt{1-x^2}}$$
I have to conside the signs of the function and then divide it in $f$ and $-f$ because I can't take all the values given by the parametrization with this, but the $f$ I got is still different from the answer I have, being the answer the $f$ I wrote above and the one I got $f(x)=\log\left( \frac{x}{1+\sqrt{1-x^2}}\right) - \sqrt{1-x^2}$.
(2) Proving the pseudosphere is orientable. Here I'm clueless. I know that I have to find a vector field of normal unitary vectors to the pseudosphere, but I don't know what to do in order to define an orientation.
HINTS:
(1) Parametrize this as a surface of revolution $\mathbf r(t,u)$.
(2) Once you check this is a regular parametrization, won't $\dfrac{\partial\mathbf r}{\partial t}\times \dfrac{\partial\mathbf r}{\partial u}$ give you a continuous, nowhere-zero normal vector field?