Question

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

• A. $\frac{2{\sin }^{2}2\theta }{\sin \theta }$
• B. $8{\cos }^3\theta -4\cos \theta +6$
• C. $\frac{2\sin 2\theta }{\sin \theta }$
• D. $8{\cos }^3\theta +4\cos \theta +6$

Answer 1

Answer: (B)

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