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# If $C=2\cos \theta$ then the value of the determinant $\Delta=\begin{vmatrix}c&1&0\\1&c&1\\6&1&c\end{vmatrix}$ is

$\begin{array}{1 1}(a)\;\large\frac{2\sin^2\theta}{\sin \theta}&(b)\;8\cos^3\theta-4\cos\theta+6\\(c)\;\large\frac{2\sin 2\theta}{\sin \theta}&(d)\;8\cos^3\theta+4\cos \theta+6\end{array}$

Can you answer this question?

Given :
$\Delta=\begin{vmatrix}c&1&0\\1&c&1\\6&1&c\end{vmatrix}$
$\quad=c(c^2-1)-1(c-6)$
$\Delta=2\cos \theta(4\cos^2\theta-1)-(2\cos \theta-6)$
Given :c=$2\cos\theta$
$\Delta=8\cos^3\theta-4\cos\theta+6$
Hence (b) is the correct answer.
answered Nov 22, 2013