If $\alpha and\beta$ are 2 distinct roots of equation $a\cos \theta +b\sin \theta =c$, then $\cos (\alpha +\beta )=$

$b\sin \theta =-c-a\cos \theta$

$b^2(1-\cos 2\theta )=c^2+a^2\cos 2\theta -2ac\cos \theta$

$\Rightarrow (a^2+b^2)\cos 2\theta -2ac\cos \theta +(c^2-b^2)=0$

$\cos \alpha \cdot \cos \beta =\frac{c^2-b^2}{a^2+b^2}\to \left(1\right)$

$a^2(1-\sin 2\theta )=c^2+b^2\sin 2\theta -2bc\sin \theta$

$(a^2+b^2)\sin 2\theta -2bc\sin \theta +(c^2-a^2)=0$

$\sin \alpha \cdot \sin \beta =\frac{c^2-a^2}{a^2+b^2}\to \left(2\right)$

$\cos (\alpha +\beta )=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta =\frac{a^2-b^2}{a^2+b^2}$