Question

If $C=2\cos \theta$ , then the value of the determinant $△=\left|\begin{array}{ccc}C& 1& 0\\ 1& C& 1\\ 6& 1& C\end{array}\right|$ is :

• A. $\frac{\sin 4\theta }{\sin \theta }$
• B. $\frac{2{\sin }^2\theta }{\sin \theta }$
• C. $4{\cos }^2\theta (2\cos \theta -1)$
• D. None of these above

Answer 1

The correct option is D None of these above

Let $△=\left|\begin{array}{ccc}C& 1& 0\\ 1& C& 1\\ 6& 1& C\end{array}\right|=C(C^2-1)-1(C-6)=C^3-2C+6$
Put $C=2\cos \theta$ we get
$△=(2\cos \theta {)}^3-2(2\cos \theta )+6$
$=8{\cos }^3\theta -4\cos \theta +6$