find the equation of the line passing through the points (a cos alpha, asin alpha) and (acos beta, a sin beta)

Given points are $\left(a\cos \alpha ,a\sin \alpha \right)$ and $\left(a\cos \beta ,a\sin \beta \right)$
Suppose $y=mx+c$ be the equation of line passing through the two given points.
So slope of the line; $m=\frac{\left(a\sin \beta -a\sin \alpha \right)}{\left(a\cos \beta -a\cos \alpha \right)}$
=$$\begin{gathered} m=\left(\frac{\sin \beta -\sin \alpha }{\cos \beta -\cos \alpha }\right) \\ =\frac{2\sin \left({\displaystyle \frac{\beta -\alpha }{2}}\right).\cos \left({\displaystyle \frac{\beta +\alpha }{2}}\right)}{2\sin \left({\displaystyle \frac{\beta +\alpha }{2}}\right).\sin \left({\displaystyle \frac{\beta -\alpha }{2}}\right)} \\ =\frac{\cos \left({\displaystyle \frac{\beta +\alpha }{2}}\right)}{\sin \left({\displaystyle \frac{\beta +\alpha }{2}}\right)} \\ =cot\left(\frac{\beta +\alpha }{2}\right) \end{gathered}$$
So equation of line is $y=cot\left(\frac{\beta +\alpha }{2}\right)x+c$
As this line passes through $\left(a\cos \alpha ,a\sin \alpha \right)$; putting $x=a\cos \alpha$ and $y=a\sin \alpha$ in the above equation we get;
$$\begin{gathered} a\sin \alpha =cot\left(\frac{\beta +\alpha }{2}\right)a\cos \alpha +c \\ \Rightarrow c=a\sin \alpha -a\cos \alpha cot\left(\frac{\beta +\alpha }{2}\right)=a\left(\sin \alpha -\cos \alpha cot\left(\frac{\beta +\alpha }{2}\right)\right) \end{gathered}$$

So the equation of required line is $y=cot\left(\frac{\beta +\alpha }{2}\right)x+a\left(\sin \alpha -\cos \alpha cot\left(\frac{\beta +\alpha }{2}\right)\right)$