Q. If $C = 2\cos\theta$, then the value of the determinant $\mathrm{\Delta} = \left| \begin{matrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{matrix} \right|$ is
A) $\frac{\sin{4\theta}}{\sin\theta}$
B) $\frac{2\sin^{2}{2\theta}}{sin\theta}$
C) $4\cos^{2}{\theta(2\cos{\theta - 1)}}$
D) None of these
Explanation
(d)
Given that $C = 2\cos\theta$
and $\mathrm{\Delta} = \left| \begin{matrix}
C & 1 & 0 \\
1 & C & 1 \\
6 & 1 & C
\end{matrix} \right| = C\left( C^{2} - 1 \right) - 1(C - 6)$
$\mathrm{\Delta} = 2\cos{\theta(4\cos^{2}{\theta - 1) - (2\cos{\theta - 6)}}}$ $(\because C = 2\cos{\theta)}$
$\Rightarrow \ \mathrm{\Delta} = 8\cos^{3}{\theta - 4\cos{\theta + 6}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.