Q. The value of the determinant $\left| \begin{matrix} a^{2} & a & 1 \\ \cos{(nx)} & \cos{(n + 1)x} & \cos{(n + 2)x} \\ \sin{(nx)} & \sin{(n + 1)x} & \sin{(n + 2)x} \end{matrix} \right|$ is independent of

A) $n$
B) $a$
C) $x$
D) None of these
Explanation

(a)
Le t$\mathrm{\Delta} = \left| \begin{matrix}
a^{2} & a & 1 \\
\cos{(nx)} & \cos{(n + 1)x} & \cos{(n + 2)x} \\
\sin{(nx)} & \sin{(n + 1)x} & \sin{(n + 2)x}
\end{matrix} \right|$
Since, $\cos{(nx)} + \cos{(n + 2)x} = 2\cos{(n + 1)x}\cos x$
and $\sin{(nx)} + \sin{(n + 2)x} = 2\sin{(n + 1)x}\cos x$
Applying $C_{1} \rightarrow C_{1} - 2\cos{x \bullet C_{2} + C_{3}}$
$\therefore\ \mathrm{\Delta} = \left| \begin{matrix}
a^{2} - 2a\cos{x + 1} & a & 1 \\
0 & \cos{(n + 1)x} & \cos{(n + 2)x} \\
0 & \sin{(n + 1)x} & \sin{(n + 2)x}
\end{matrix} \right|$
$= \left( a^{2} - 2a\cos{x + 1} \right)\lbrack\cos{(n + 1)x}\sin{(n + 2)x} - \cos{(n + 2)x}\ \sin{(n + 1)x}\rbrack$
$= \left( a^{2} - 2a\cos{x + 1} \right)\sin x$
$\therefore\ \mathrm{\Delta}$ is independent of $n$.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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