Question

The value of the determinant
$\left|\begin{array}{ccc}1& a& a^2\\ cos(n-1)x& cosnx& cos(n+1)x\\ sin(n-1)x& sinnx& sin(n+1)x\end{array}\right|(a\ne 1)$ is zero if

• A. $sinx=0$
• B. $cosx=0$
• C. $a=0$
• D. $cosx=\frac{1+a^2}{2a}$

Answer 1

The correct option is A $sinx=0$

Applying $C_1\to C_1+C_3-2cosx{C}_2$, the given determinant is equal to

$\left|\begin{array}{ccc}1+a^2-2acosx& a& a^2\\ 0& cosnx& cos(n+1)x\\ 0& sinnx& sin(n+1)x\end{array}\right|$

$(1+a^2-2acosx).\left[cosnxsin\right(n+1)x-sinnxcos(n+1\left)x\right]$

$=(1+a^2-2acosx)sinx(n+1-n)x$

$=(1+a^2-2acosx)sinx$

which is zero if $sinx=0$ or $cosx=\frac{(1+a^2)}{2a}$.

As $a\ne 1$,

$\frac{(1+a^2)}{2a}>1$

Therefore, $cosx=\frac{(1+a^2)}{2a}$ is not possible.