Question

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

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

Answer 1

The correct option is D 0

$\left|\begin{array}{ccc}1& cos(\beta -\alpha )& cos\alpha \\ cos(\alpha -\beta )& 1& cos\beta \\ cos\alpha & cos\beta & 1\end{array}\right|$

$=\left|\begin{array}{ccc}1& \cos \left(\alpha -\beta \right)& \cos \alpha \\ \cos \left(\alpha -\beta \right)& 1& \cos \beta \\ \cos \alpha & \cos \beta & 1\end{array}\right|$

$=\left|\begin{array}{ccc}1& \cos \alpha \cos \beta & \cos \alpha \\ \cos \alpha \cos \beta & 1& \cos \beta \\ \cos \alpha & \cos \beta & 1\end{array}\right|$

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

$=\left|\begin{array}{ccc}{\sin }^2\alpha & \sin \alpha .\sin \beta & \cos \alpha \\ \sin \alpha .\sin \beta & {\sin }^2\beta & \cos \beta \\ 0& 0& 1\end{array}\right|$

$={\sin }^2\alpha .{\sin }^2\beta -{\sin }^2\alpha .{\sin }^2\beta$

$=0$