Q. If $c = 2\cos{\theta,\ }\$then the value of the determinant $\mathrm{\Delta} = \left| \begin{array}{r} c\ \ 1\ \ 0 \\ 1\ \ c\ \ 1 \\ 6\ \ 1\ \ c \end{array} \right|$ is
A) $\frac{sin4\theta\ }{\sin\theta}$
B) $\frac{2\sin^{2}2\theta\ }{\sin\theta}$
C) $4\cos^{2}{\theta(2\cos{\theta - 1)}}$
D) None of these
Explanation
(d)
Let $\mathrm{\Delta} = \left| \begin{array}{r}
c\ \ 1\ \ 0 \\
1\ \ c\ \ 1 \\
6\ \ 1\ \ c
\end{array} \right| = c\left( c^{2} - 1 \right) - 1(c - 6)$
$= 8\cos^{3}{\theta - 4\cos{\theta + 6}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.