Question

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

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