Cosine of the sum of two solutions of trigonometric equation $a\cos \theta + b\sin \theta = c$ Question:
If $\alpha$ and $\beta$ are the solutions of $a\cos \theta + b\sin \theta = c$, then show that:
$$\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2}$$
No idea how to even approach the problem. I tried taking two equations, by substituting $\alpha$ and $\beta$ in place of $\theta$ in the equation and manipulating them, but that didn't get me anywhere. Please help!

We have $\displaystyle a\cos\theta=c-b\sin\theta,$
Squaring we get,
$\displaystyle(c-b\sin\theta)^2=(a\cos\theta)^2=a^2(1-\sin^2\theta)$
$\displaystyle\iff (a^2+b^2)\sin^2\theta-2bc\sin\theta+c^2-a^2=0$
So, $\displaystyle\sin\alpha\sin\beta=\dfrac{c^2-a^2}{a^2+b^2}$
Similarly find $\displaystyle\cos\alpha\cos\beta$ by squaring $\displaystyle b\sin\theta=c-a\cos\theta$
Finally use $\displaystyle\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
