If minor of three-one element `(i.e.M_(31))` in the determinant `[{:(,0,1,sec alpha),(,tan alpha,-sec alpha,tan alpha),(,1,0,1)]:}` is 1 then find the value of `alpha (0 le alpha le pi).`

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the minor of the element \( M_{31} \) in the given determinant is equal to 1. The determinant is given as:

\[ \begin{vmatrix} 0 & 1 & \sec \alpha \\ \tan \alpha & -\sec \alpha & \tan \alpha \\ 1 & 0 & 1 \end{vmatrix} \]

### Step 1: Calculate the Minor \( M_{31} \)

The minor \( M_{31} \) is obtained by deleting the third row and the first column from the determinant. The remaining 2x2 determinant is:

\[ \begin{vmatrix} 1 & \sec \alpha \\ \tan \alpha & \tan \alpha \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinant

The determinant of a 2x2 matrix is calculated as follows:

\[ \text{Determinant} = (1)(\tan \alpha) - (\sec \alpha)(\tan \alpha) \]

This simplifies to:

\[ \tan \alpha - \sec \alpha \tan \alpha \]

### Step 3: Factor Out \( \tan \alpha \)

We can factor out \( \tan \alpha \):

\[ \tan \alpha (1 - \sec \alpha) \]

### Step 4: Set the Minor Equal to 1

According to the problem, we have:

\[ M_{31} = \tan \alpha (1 - \sec \alpha) = 1 \]

### Step 5: Substitute \( \sec \alpha \)

Recall that \( \sec \alpha = \frac{1}{\cos \alpha} \). Thus, we can rewrite the equation:

\[ \tan \alpha \left(1 - \frac{1}{\cos \alpha}\right) = 1 \]

### Step 6: Simplify the Equation

Using the identity \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \), we can rewrite the equation as:

\[ \frac{\sin \alpha}{\cos \alpha} \left(1 - \frac{1}{\cos \alpha}\right) = 1 \]

This simplifies to:

\[ \frac{\sin \alpha}{\cos \alpha} \cdot \frac{\cos \alpha - 1}{\cos \alpha} = 1 \]

### Step 7: Cross Multiply

Cross multiplying gives us:

\[ \sin \alpha (\cos \alpha - 1) = \cos \alpha \]

### Step 8: Rearranging the Equation

Rearranging the equation leads to:

\[ \sin \alpha \cos \alpha - \sin \alpha = \cos \alpha \]

### Step 9: Move All Terms to One Side

This can be rearranged to:

\[ \sin \alpha \cos \alpha - \cos \alpha - \sin \alpha = 0 \]

### Step 10: Factor the Equation

Factoring out \( \cos \alpha \):

\[ \cos \alpha (\sin \alpha - 1) - \sin \alpha = 0 \]

### Step 11: Solve for \( \alpha \)

Setting each factor to zero gives us two cases:

1. \( \cos \alpha = 0 \)
2. \( \sin \alpha - 1 = 0 \)

From \( \cos \alpha = 0 \), we find:

\[ \alpha = \frac{\pi}{2} \]

From \( \sin \alpha - 1 = 0 \), we find:

\[ \sin \alpha = 1 \implies \alpha = \frac{\pi}{2} \]

### Step 12: Check the Range

The values of \( \alpha \) must be in the range \( [0, \pi] \). The solutions are:

\[ \alpha = 0, \frac{\pi}{2}, \text{ and } \frac{3\pi}{4} \]

### Final Answer

Thus, the values of \( \alpha \) are:

\[ \alpha = 0, \frac{\pi}{2}, \text{ and } \frac{3\pi}{4} \]