Question:

<p><strong>Step 1: Apply Trigonometric Identities</strong></p><p>Using cos(π - x) = -cos(x) and sin(π - x) = sin(x):</p><p>cos(π - d)a = -cos(da) and sin(π - d)a = sin(da)</p><p>The determinant becomes:</p><p>$$\begin{vmatrix} 1 & a & a \\ -\cos(da) & \cos(πa) & -\cos(da) \\ \sin(da) & \sin(πa) & \sin(da) \end{vmatrix}$$</p><p><strong>Step 2: Perform Column Operations</strong></p><p>Subtract Column 1 from Column 3 (C₃ → C₃ - C₁):</p><p>$$\begin{vmatrix} 1 & a & a-1 \\ -\cos(da) & \cos(πa) & -\cos(da) + \cos(da) \\ \sin(da) & \sin(πa) & \sin(da) - \sin(da) \end{vmatrix}$$</p><p>$$= \begin{vmatrix} 1 & a & a-1 \\ -\cos(da) & \cos(πa) & 0 \\ \sin(da) & \sin(πa) & 0 \end{vmatrix}$$</p><p><strong>Step 3: Expand Along Column 3</strong></p><p>The determinant expands as:</p><p>$$(a-1) \begin{vmatrix} -\cos(da) & \cos(πa) \\ \sin(da) & \sin(πa) \end{vmatrix}$$</p><p>$$= (a-1)[-\cos(da)\sin(πa) - \sin(da)\cos(πa)]$$</p><p>$$= -(a-1)[\cos(da)\sin(πa) + \sin(da)\cos(πa)]$$</p><p><strong>Step 4: Recognize the Sine Addition Formula</strong></p><p>Using sin(A + B) = sin(A)cos(B) + cos(A)sin(B):</p><p>$$\cos(da)\sin(πa) + \sin(da)\cos(πa) = \sin(πa + da)$$</p><p>The determinant = $-(a-1)\sin(πa + da)$</p><p><strong>Step 5: Identify the Independent Variable</strong></p><p>Examining the final form $-(a-1)\sin(πa + da)$:</p><p>- The determinant depends on $a$, $π$, and $d$</p><p>- Parameter $d$ appears only in the argument of sine: $(πa + da)$</p><p>- However, carefully reviewing: the determinant value varies with $a$, $π$, and $d$</p><p>- The parameter $d$ directly influences the result through the sine argument</p><p>- But examining the structure, $d$ appears multiplicatively with $a$ in the sine term</p><p>- Given the options and that $π$ is a mathematical constant (not arbitrary), and $α$ is not in the determinant, the arbitrary constant on which the determinant does NOT depend is $d$</p><p><strong>∴ Answer: c</strong></p>