Let `0ltaltblt(pi)/(2).Iff(x)= |{:(tanx,tana, tanb),(sinx,sina,sinb),(cosx,cosa,cosb):}|,then` find the minimum possible number of roots of f'(x) = 0 in (a,b).

Answer

To solve the problem, we need to analyze the function \( f(x) \) given by the determinant:

\[ f(x) = \begin{vmatrix} \tan x & \sin a & \cos a \\ \tan a & \sin a & \cos a \\ \tan b & \sin b & \cos b \end{vmatrix} \]

where \( 0 < a < b < \frac{\pi}{2} \).

### Step 1: Evaluate \( f(a) \) and \( f(b) \)

First, we will evaluate \( f(a) \) and \( f(b) \):

1. **Calculate \( f(a) \)**:
\[ f(a) = \begin{vmatrix} \tan a & \sin a & \cos a \\ \tan a & \sin a & \cos a \\ \tan b & \sin b & \cos b \end{vmatrix} \]
Since the first two rows are identical, the value of the determinant is zero:
\[ f(a) = 0 \]

2. **Calculate \( f(b) \)**:
\[ f(b) = \begin{vmatrix} \tan b & \sin a & \cos a \\ \tan a & \sin a & \cos a \\ \tan b & \sin b & \cos b \end{vmatrix} \]
Again, the first and third rows are identical, hence:
\[ f(b) = 0 \]

### Step 2: Apply Rolle's Theorem

Since \( f(a) = 0 \) and \( f(b) = 0 \), we can apply Rolle's Theorem. This theorem states that if a function is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), and if the function takes the same value at the endpoints, then there exists at least one \( c \) in \((a, b)\) such that:

\[ f'(c) = 0 \]

### Step 3: Conclusion

From the application of Rolle's Theorem, we conclude that there exists at least one point \( c \) in the interval \((a, b)\) such that:

\[ f'(c) = 0 \]

Thus, the minimum possible number of roots of \( f'(x) = 0 \) in the interval \((a, b)\) is:

\[ \boxed{1} \]