Question

If $\alpha$ and $\beta$ are the solution of $a$ $\cos \theta +b\sin \theta =c$, then show that :

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

Answer 1

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