Why is this arctangent a line on the interval $t = (-1, 1)$
$$ \arctan \left(\frac{\cos(\frac{\alpha}{2}t)-\cos(\frac{\alpha}{2})}{\sin(\frac{\alpha}{2}t)-t*\sin(\frac{\alpha}{2})}\right) $$
I don't know of a trigonometric identity that simplifies this.
For context, I believe it is the angle between a point $x = \frac{c}{2}t$ on the chord of a circular segment with central angle $\alpha$ and the point on the arc $\theta = \frac{\alpha}{2}t$. I'm attempting to describe a circular arc/circular segment in terms of its curvature or degree of curvature rather than its center and radius.
Assuming that additional $t$ in the end of your denominator is a typo, let
$$f(t) = \arctan\left( \frac{\cos(at)-\cos(a)}{\sin(at)-\sin(a)} \right)\tag 1$$
where $a=\alpha/2$. The addition theorems for numerator and denominator of the argument of arctan yield
$$\cos(at)-\cos(a) = -2\sin\frac{at+a}2 \sin\frac{at-a}2 \tag3$$ $$\sin(at)-\sin(a) = 2\cos\frac{at+a}2 \sin\frac{at-a}2 \tag4$$
so that the quotient (3)/(4) is just tan:
$$\begin{align} f(t) &= \arctan\left(-\tan\frac{at+a}2\right)\\ &= -\arctan\left(\tan\frac{at+a}2\right)\\ &= -\frac{at+a}2 \bmod \pi\\ &= -\frac\alpha 4 (t+1) \bmod \pi \tag 5\\ \end{align}$$ provided we chose the remainder $\bmod \pi \in[-\pi/2,\pi/2]$.