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(A) Let the determinant be $\Delta$. We perform the column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_2$ is not ideal here. Instead,let us ex

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$n$

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(A) Let the determinant be $\Delta$. We perform the column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_2$ is not ideal here. Instead,let us expand along the first row:$\Delta = a^2 [\cos(n+1)x \sin(n+2)x - \sin(n+1)x \cos(n+2)x] - a [\cos(nx) \sin(n+2)x - \sin(nx) \cos(n+2)x] + 1 [\cos(nx) \sin(n+1)x - \sin(nx) \cos(n+1)x]$Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we get:$\Delta = a^2 \sin((n+2)x - (n+1)x) - a \sin((n+2)x - nx) + \sin((n+1)x - nx)$$\Delta = a^2 \sin(x) - a \sin(2x) + \sin(x)$$\Delta = a^2 \sin(x) - a(2 \sin x \cos x) + \sin(x)$$\Delta = \sin(x) [a^2 - 2a \cos x + 1]$Since the expression $\Delta = \sin(x) (a^2 - 2a \cos x + 1)$ still contains $a$ and $x$,we re-evaluate the question. The determinant value depends on $a$ and $x$. However,if we look at the structure,the value is independent of $n$ because the $n$ terms cancel out during the subtraction of angles in the sine identity.

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

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