The determinant `Delta=|(a^2,a,1),(cos(nx),cos (n + 1) x,cos(n+2) x),(sin(nx),sin (n +1)x,sin (n + 2) x)|` is independent of

Options

A. n
B. a
C. x
D. none of these

Explanation

To solve the determinant

\[ \Delta = \begin{vmatrix} a^2 & a & 1 \\ \cos(nx) & \cos((n+1)x) & \cos((n+2)x) \\ \sin(nx) & \sin((n+1)x) & \sin((n+2)x) \end{vmatrix} \]

we will expand it and analyze the result to determine its dependency on \( n \).

### Step 1: Expand the Determinant
We can use the formula for the determinant of a 3x3 matrix:

\[ \Delta = a_1 \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} \]

Where \( a_1, a_2, a_3 \) are the elements of the first row, and \( b_i, c_i \) are the elements of the second and third rows respectively.

### Step 2: Calculate the Minors
We will calculate the minors:

1. For \( a_1 = a^2 \):
\[ \begin{vmatrix} \cos((n+1)x) & \cos((n+2)x) \\ \sin((n+1)x) & \sin((n+2)x) \end{vmatrix} = \cos((n+1)x) \sin((n+2)x) - \sin((n+1)x) \cos((n+2)x) \]
This simplifies to:
\[ \sin((n+2)x - (n+1)x) = \sin(x) \]

2. For \( a_2 = a \):
\[ -a \begin{vmatrix} \cos(nx) & \cos((n+2)x) \\ \sin(nx) & \sin((n+2)x) \end{vmatrix} = -a (\cos(nx) \sin((n+2)x) - \sin(nx) \cos((n+2)x)) = -a \sin((n+2)x - nx) = -a \sin(x) \]

3. For \( a_3 = 1 \):
\[ \begin{vmatrix} \cos(nx) & \cos((n+1)x) \\ \sin(nx) & \sin((n+1)x) \end{vmatrix} = \cos(nx) \sin((n+1)x) - \sin(nx) \cos((n+1)x) = \sin((n+1)x - nx) = \sin(x) \]

### Step 3: Combine the Results
Now we can combine the results from the minors:

\[ \Delta = a^2 \sin(x) - a (-\sin(x)) + 1 \sin(x) \]
\[ = a^2 \sin(x) + a \sin(x) + \sin(x) \]
\[ = \sin(x)(a^2 + a + 1) \]

### Step 4: Analyze Dependency
From the final expression \( \Delta = \sin(x)(a^2 + a + 1) \), we can see that:

- The determinant \( \Delta \) depends on \( a \) and \( x \).
- It does not depend on \( n \).

### Conclusion
Thus, the determinant \( \Delta \) is independent of \( n \).