If $\alpha ,\beta$  are solution of equation $a\cos \theta +b\sin \theta =c$ Then show that

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

ii) $\cos \alpha +\cos \beta =\frac{2ac}{a^2+b^2}$

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

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

Given $a\cos \theta +b\sin \theta =c$

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

Squaring on both sides $a^2{\cos }^2\theta =(c-b\sin \theta {)}^2$

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

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

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

$\Rightarrow \left(a^2+b^2\right){\sin }^2\theta -2bc\sin \theta +c^2-a^2=0$ $\to$ (1)

Given $\alpha ,\beta$ are solutions of equation (1)

Let $\sin \alpha ,\sin \beta$   be the roots of (1)

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

ii) $\cos \alpha +\cos \beta =\frac{2ac}{a^2+b^2}$

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

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

again consider $a\cos \theta +b\sin \theta =c$

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

Squaring on both sides $b^2{\sin }^2\theta =(c-a\cos \theta {)}^2$

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

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

$\Rightarrow c^2+a^{2}c{\cos }^2\theta -2ac\cos \theta +\mid b^2{\cos }^2\theta -b^2=0$

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

Given $\alpha ,\beta$ are solutions of equation - (2)

Let $\cos \alpha ,\cos \beta$  be the roots of equation - (2)