Question

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

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

Answer 1

Answer: (A)

Given determinant
$\left|\begin{array}{ccc}1& a& a^2\\ cos\left(n-1\right)x& cosnx& cos\left(n+1\right)x\\ sin\left(n+1\right)x& sinnx& sin\left(n+1\right)x\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1+a^2-2acosx& a& a^2\\ 0& cosnx& cos\left(n+1\right)x\\ 0& sinnx& sin\left(n+1\right)x\end{array}\right|=0$
By applying $C_1\to C_1+C_3-2cosx{C}_2$
By expanding
$\left(1+a^2-2acosx\right)\left[cosnxsin\left(n+1\right)x-sinnxcos\left(n+1\right)x\right]=0$
Now, $\left(1+a^2-2acosx\right)sin\left(n+1-n\right)x=0$
$\Rightarrow \left(1+a^2-2acosx\right)sinx=0$
$sinx=0$ or $cosx=\frac{1+a^2}{2a}$
As a $\ne 1\therefore \left(\frac{1+a^2}{2a}\right)>1$
$\Rightarrow cosx>1$ It is not possible.
$\therefore sinx=0$