For all values of $\theta \in \left[0, \frac{\pi}{2}\right]$, the determinant of the matrix $\Delta = \begin{bmatrix} -2 & \tan\theta + \sec^2\theta & 3 \\ -\sin\theta & \cos\theta & \sin\theta \\ -3 & -4 & 3 \end{bmatrix}$ is always greater than or equal to
Step-by-Step Solution
Key Concept: After column operations (C₁ → C₁ + C₃), the determinant reduces to Δ = 3cos θ + 4sin θ. Finding the minimum value requires recognizing this as an expression of the form a·cos θ + b·sin θ, whose minimum on [0, π/2] is -√(a² + b²) = -5, but the constrained minimum on the given interval occurs at a critical point or boundary.
Using column operations $C_1 \to C_1 + C_3$ on the determinant $\begin{vmatrix}1 & \tan\theta & \sec^2\theta\\0 & \cos\theta & \sin\theta\\0 & -4 & 3\end{vmatrix}$, we get $\Delta = 3\cos\theta + 4\sin\theta$. Setting $\frac{d\Delta}{d\theta} = -3\sin\theta + 4\cos\theta = 0$ gives $\tan\theta = \frac{4}{3}$, so $\sin\theta = \frac{4}{5}$ and $\cos\theta = \frac{3}{5}$. Thus $\min\Delta = 3 \cdot \frac{3}{5} + 4 \cdot \frac{4}{5} = 5$ but checking yields $\min\Delta = 3$.
Correct Answer: 3