Are these charts on the circle compatibly oriented? I've tried a few methods but I can't seem to work this one out.
Consider the charts
$$f(s) = (\cos s, \sin s) \in \mathbb{R}^2$$
for $-\pi < s < \pi$ and 
$$g(t)=(\frac{2t}{t^2 + 1}, \frac{t^2 - 1}{t^2 + 1})$$
for $t \in \mathbb{R}$. Are these charts on the circle compatibly oriented? 

We will use the identities
$$ \sin(2 \theta) = \frac{2 \tan(\theta)}{1 + \tan^2(\theta)}, \;\;\; \cos(2 \theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}. $$
Making the change of variable $t = \tan(\theta)$, we define a chart by
$$ \psi(\theta) = g(\tan(\theta)) = (\sin(2\theta), -\cos(2\theta)) : (-\frac{\pi}{2}, \frac{\pi}{2}) \rightarrow \mathbb{R}^2. $$
Drawing the situation, we see that both $f$ and $\psi$ trace the circle counter-clockwise as we go from lower values of $s$ (or $\theta$) to higher values. This implies that the charts are consistently oriented. To show this rigorously, write
$$ \cos(s) = \sin(2\theta) = \cos(\frac{\pi}{2} - 2\theta). $$
Taking into the account the domains of $\theta$ and $s$, we have then
$$ 2\theta = \left\{
     \begin{array}{lr}
       s + \frac{\pi}{2}, \;\;\; -\pi < s < \frac{\pi}{2}\\
       s - \frac{3\pi}{2},\;\;\;\; \frac{\pi}{2} \leq s < \pi.
     \end{array}
   \right.$$
(Why is there a discontinuity in the description?)
We see that the coordinate transformation from $\theta$ to $s$ has positive Jacobian, and as this is also true for the transformation from $t$ to $\theta$, the charts are indeed consistently oriented.
