Gaussian curvature of the pseudosphere I am asked to show that the pseudosphere has Gaussian Curvature $-1$ at all points. So I parametrized the tractrix as:
\begin{equation}
\alpha(t) = (\sin{t},\cos{t}+\log{\tan{\frac{t}{2}}})
\end{equation}
So the surface of revolution would be:
\begin{equation}
x(\theta,t)=(\sin{t}\cos{\theta},\sin{t}\sin{\theta},\cos{t}+\log{\tan{\frac{t}{2}}})
\end{equation}
By calculating the first and second fundamental forms we have that $F=f=0$ and $E=(\sin{t})^2,G=(\cot{t})^2,e=(\cos{t})^2$ and $g=-(\cot{t})^2$. So the Gaussian curvature is:
\begin{equation}
K = \frac{eg}{EG} = -(\cot{t})^2
\end{equation}
What am I doing wrong?
 A: We have
\begin{align}
x_\theta &= (-\sin t\sin\theta, \sin t \cos \theta, 0) = \sin t(-\sin\theta, \cos\theta, 0) \\
x_t &= (\cos t \cos\theta, \cos t \cos\theta, -\sin t + \frac{\sec^2(t/2)}{2\tan(t/2)}) = \cos t(\cos\theta, \sin\theta, \cot t),
\end{align}
so
\begin{align}
E&= \langle x_\theta, x_\theta \rangle = \sin^2 t\\
F&= 0 \quad\text{[surface of revolution]}\\
G&= \langle x_t, x_t \rangle = \cos^2 t (1+\cot^2t) = \cot^2t.
\end{align}
So far so good. Now $x_\theta \times x_t$ is parallel to
$$ (-\sin\theta, \cos\theta, 0) \times(\cos\theta,\sin\theta,\cot t) = (\cos\theta\cot t, \sin\theta\cot t, -1) $$
so we can (after normalizing to make it a unit vector, i.e. multiplying by $\sin t$) choose the Gauss map to be
$$ N = (\cos\theta\cos t, \sin\theta\cos t, -\sin t). $$
This gives
\begin{align}
N_\theta &= (-\sin\theta\cos t, \cos\theta\cos t, 0)\\
N_t &= (-\cos\theta\sin t, -\sin\theta\sin t, -\cos t),
\end{align}
so that
\begin{align}
e &= - \langle N_\theta, x_\theta \rangle = -\sin t \cos t\\
f &= 0 \quad\text{[surface of revolution]}\\
g &= - \langle N_t, x_t \rangle = -\cos t (-\sin t - \cos t \cot t) = \cot t, 
\end{align}
which gives
$$ K = \frac{eg-f^2}{EG-F^2} = \frac{-\cos^2t}{\sin^2t\cot^2t} = -1. $$
