# If $C=2\cos \theta$ then the value of the determinant $4\Delta=\begin{vmatrix}c&1&0\\1&c&1\\0&1&c\end{vmatrix}$ is

$\begin{array}{1 1}(A)\;\large\frac{\sin 4\theta}{\sin \theta}&(B)\;\large\frac{2\sin ^2\theta}{\sin \theta}\\(C)\;4\cos^2\theta(2\cos \theta-1)&(D)\;\text{None }\end{array}$

$\begin{vmatrix}c&1&0\\1&c&1\\0&1&c\end{vmatrix}$
$\Rightarrow c(c^2-1)-1(c-0)$
$\Rightarrow c^3-2c$
Given $c=2\cos \theta$
$\Rightarrow 8\cos \theta-4\cos \theta$
$\Rightarrow 4\cos \theta(2\cos^2\theta-1)$
$\Rightarrow 4\cos \theta.\cos 2\theta$
$\Rightarrow \large\frac{2\times 2\cos \theta\sin \theta\cos 2\theta}{\sin \theta}$
$\Rightarrow \large\frac{4\cos \theta\sin \theta\cos 2\theta}{\sin \theta}$
$\Rightarrow \large\frac{\sin 4\theta}{\sin \theta}$
Hence (A) is the correct answer.
answered Apr 23, 2014