Question

If $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0,$ then the value of the determinant
$\left|\begin{array}{ccc}1& \cos (\beta -\alpha )& \cos \alpha \\ \cos (\beta -\alpha )& 1& \cos \beta \\ \cos \alpha & \cos \beta & 1\end{array}\right|$ , is

• A. $\sin (\alpha +\beta )$
• B. $\sin \alpha \sin \beta$
• C. $1+\cos (\alpha +\beta )$
• D. $0$

Answer 1

The correct option is D $0$

$\begin{array}{c}=\left|\begin{array}{ccc}1& \cos \left(\alpha -\beta \right)& \cos \alpha \\ \cos \left(\alpha -\beta \right)& 1& \cos \beta \\ \cos \alpha & \cos \beta & 1\end{array}\right|\\ =\left|\begin{array}{ccc}1& \cos \alpha \cos \beta & \cos \alpha \\ \cos \alpha \cos \beta & 1& \cos \beta \\ \cos \alpha & \cos \beta & 1\end{array}\right|\\ =\left|\begin{array}{ccc}1-{\cos }^2\alpha & \cos \alpha \cos \beta +\sin \alpha .\sin \beta -\cos \alpha .\cos \beta & \cos \alpha \\ \cos \alpha \cos \beta \sin \alpha \sin \beta -\cos \alpha .\cos \beta & 1-{\cos }^2\beta & \cos \beta \\ \cos \alpha -\cos \beta & \cos \alpha -\cos \beta & 1\end{array}\right|\\ =\left|\begin{array}{ccc}{\sin }^2\alpha & \sin \alpha .\sin \beta & \cos \alpha \\ \sin \alpha .\sin \beta & {\sin }^2\beta & \cos \beta \\ 0& 0& 1\end{array}\right|\\ ={\sin }^2\alpha .{\sin }^2\beta -{\sin }^2\alpha .{\sin }^2\beta \\ =0\\ OptionDiscorrectanswer.\end{array}$