If $a,b,c$ are non-zero real numbers and $\alpha ,\beta$  are the solutions of the equation  $a\cos \theta +b\sin \theta =C$  Then S.T

a)  $\sin \alpha +\sin \beta =\frac{2bc}{a^2+b^2}$

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

 $a\cos \theta +b\sin \theta =C$ $\Rightarrow a\cos \theta =c-b\sin \theta$

S.O.B.S $\Rightarrow (a\cos \theta {)}^2=(c-b\sin \theta {)}^2$

$\Rightarrow a^2{\cos }^2\theta =c^2+b^2{\sin }^2\theta -2bc\sin \theta$

$\Rightarrow a^2\left(1-{\sin }^2\theta \right)=c^2-2bc\sin \theta +b^2{\sin }^2\theta$

$\Rightarrow \left(a^2+b^2\right){\sin }^2\theta -2bc\sin \theta +\left(c^2-a^2\right)=0$

This is a quadratic equation in $\sin \theta$ whose roots  are $\sin \alpha$ and  $\sin \beta$  (given that $\alpha ,\beta$. are two
solutions of $\theta$) Therefore

$\sin \alpha +\sin \beta$ = sum of roots = $\frac{2bc}{a^2+b^2}$

$\sin \alpha .\sin \beta$ = Product of roots $=\frac{c^2-a^2}{a^2+b^2}$