`|(sin^(2) x,cos^(2) x,1),(cos^(2) x,sin^(2) x,1),(- 10,12,2)| =`

Options

A. 0
B. `12 cos^(2) x -10 sin^(2)x`
C. `12 sin^(2) x - 10 cos^(2) x - 2`
D. `10 sin 2x`

Answer

To solve the determinant

\[ D = \begin{vmatrix} \sin^2 x & \cos^2 x & 1 \\ \cos^2 x & \sin^2 x & 1 \\ -10 & 12 & 2 \end{vmatrix} \]

we will apply some properties of determinants and trigonometric identities.

### Step 1: Apply Column Operations
We will perform the operation \( C_1 \rightarrow C_1 + C_2 \). This means we will add the second column to the first column.

\[ D = \begin{vmatrix} \sin^2 x + \cos^2 x & \cos^2 x & 1 \\ \cos^2 x + \sin^2 x & \sin^2 x & 1 \\ -10 + 12 & 12 & 2 \end{vmatrix} \]

### Step 2: Simplify Using Trigonometric Identity
Using the trigonometric identity \( \sin^2 x + \cos^2 x = 1 \):

\[ D = \begin{vmatrix} 1 & \cos^2 x & 1 \\ 1 & \sin^2 x & 1 \\ 2 & 12 & 2 \end{vmatrix} \]

### Step 3: Apply Another Column Operation
Now, we will perform the operation \( C_1 \rightarrow C_1 - C_3 \):

\[ D = \begin{vmatrix} 1 - 1 & \cos^2 x & 1 \\ 1 - 1 & \sin^2 x & 1 \\ 2 - 2 & 12 & 2 \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} 0 & \cos^2 x & 1 \\ 0 & \sin^2 x & 1 \\ 0 & 12 & 2 \end{vmatrix} \]

### Step 4: Identify Zero Row
Now we can see that the first column has all zeros in the first two rows.

### Step 5: Conclusion
Since there is a row of zeros in the determinant, the value of the determinant is:

\[ D = 0 \]

Thus, the final answer is:

\[ \boxed{0} \]

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