Question

If $\alpha$ and $\beta$ are the solution of $a$ $\cos \theta +b\sin \theta =c$, then show that : $\cos (\alpha -\beta )=\frac{2c^2-(a^2+b^2)}{a^2+b^2}$

Answer 1

$\because cos(\alpha -\beta )=cos\alpha cos\beta +sin\alpha sin\beta ⟶\left(1\right)also,acos\theta +bsin\theta =c\Rightarrow acos\theta =c-bsin\theta \Rightarrow a^2{cos}^2\theta =c^{2}cbsin\theta +b^2{sin}^2\theta \left(sqauringbothside\right)\Rightarrow (a^2+b^2){sin}^2\theta -2cbsin\theta +(c^2-a^2)=0(\because {cos}^2\theta =1-{sin}^2\theta )productofroots=\frac{c}{a}\Rightarrow sin\alpha sin\beta =\frac{(c^2-a^2)}{(a^2+b^2)}⟶\left(2\right)similarly,acos\theta +bsin\theta =c\Rightarrow acos\theta -c=bsin\theta \Rightarrow a^2{cos}^2\theta -2accos\theta +c^2=b^2{sin}^2\theta \left(squaringbothside\right)\Rightarrow (a^2+b^2){cos}^2\theta -2accos\theta +(c^2-b^2)=0productofroots=\frac{c}{a}\Rightarrow cos\alpha cos\beta =\frac{(c^2-b^2)}{(a^2+b^2)}⟶\left(3\right)puttingthevalueof\left(2\right)and\left(3\right)inequation\left(1\right)\Rightarrow cos(\alpha -\beta )=\frac{(c^2-b^2)}{(a^2+b^2)}+\frac{(c^2-a^2)}{(a^2+b^2)}=\frac{2c^2-(a^2+b^2)}{(a^2+b^2)}$