If $\alpha ,\beta$ are different values of  $x$  satisfying $a\cos x+b\sin x=c$then $\tan \left(\frac{\alpha +\beta }{2}\right)=$

Change $\sin x, \cos x into \tan (x/2)$ and get quadratic in $\tan (x/2)$

$\begin{array}{rcl}a\frac{1-t^2}{1+t^2}+b\frac{2t}{1+t^2}& =& c\\ a\left(1-t^2\right)+2bt& =& c\left(1+t^2\right)\\ t^2\left(a+c\right)-2bt+c-a& =& 0\end{array}$

sum of the roots is $\frac{2b}{a+c}$ and the product of the roots is $\frac{c-a}{a+c}$

$\begin{array}{rcl}\tan \left(\frac{\alpha +\beta }{2}\right)& =& \frac{\tan \left({\displaystyle \frac{\alpha }{2}}\right)+\tan \left({\displaystyle \frac{\beta }{2}}\right)}{1-\tan \left(\frac{\alpha }{2}\right)\tan \left(\frac{\beta }{2}\right)}\\ & =& \frac{2b}{a+c-c+a}\\ & =& \frac{b}{a}\end{array}$