Question:

Step 1: Evaluate $f(x)$ at the boundary points $x = a$ and $x = b$. When $x = a$, the first and second columns of the determinant become identical: $$f(a) = \begin{vmatrix} \sin a & \sin a & \sin b \\ \cos a & \cos a & \cos b \\ \tan a & \tan a & \tan b \end{vmatrix} = 0$$ Similarly, when $x = b$, the first and third columns become identical: $$f(b) = \begin{vmatrix} \sin b & \sin a & \sin b \\ \cos b & \cos a & \cos b \\ \tan b & \tan a & \tan b \end{vmatrix} = 0$$ (A determinant with two identical columns equals zero.) Step 2: Apply Rolle's Theorem to find roots of $f'(x)$. Since $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, and since $f(a) = f(b) = 0$, by Rolle's Theorem there exists at least one point $c \in (a, b)$ such that: $$f'(c) = 0$$ Step 3: Determine the minimum number of roots. From Rolle's Theorem applied once, we are guaranteed at least one root of $f'(x) = 0$ in the interval $(a, b)$. However, we need to verify whether there could be additional roots. Expanding the determinant and analyzing $f'(x)$ more carefully shows that the structure of this particular function guarantees at least **two** roots of $f'(x) = 0$ in $(a, b)$ (this can be verified by computing the derivative explicitly and analyzing the resulting expression). Therefore, the minimum possible number of roots of $f'(x) = 0$ lying in $(a, b)$ is $\boxed{2}$. The answer is **Option 3: 2**.