if determinant `|{:( cos (theta + phi),,-sin (theta+phi),,cos 2phi),(sin theta,,cos theta,,sin phi),(-cos theta,,sintheta,,cos phi):}|` is

Options

A. non-negative
B. independent of theta
C. independent of `phi`
D. none of these

Explanation

To solve the determinant given by

\[ D = \begin{vmatrix} \cos(\theta + \phi) & -\sin(\theta + \phi) & \cos(2\phi) \\ \sin(\theta) & \cos(\theta) & \sin(\phi) \\ -\cos(\theta) & \sin(\theta) & \cos(\phi) \end{vmatrix} \]

we will use trigonometric identities and properties of determinants.

### Step 1: Use Trigonometric Identities
We can start by applying the trigonometric identities for cosine and sine of a sum:

\[ \cos(\theta + \phi) = \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi) \]
\[ \sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi) \]

Substituting these into the determinant gives:

\[ D = \begin{vmatrix} \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi) & -(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi)) & \cos(2\phi) \\ \sin(\theta) & \cos(\theta) & \sin(\phi) \\ -\cos(\theta) & \sin(\theta) & \cos(\phi) \end{vmatrix} \]

### Step 2: Simplify the Determinant
Now, we can simplify the first row using the identities we substituted:

\[ D = \begin{vmatrix} \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi) & -\sin(\theta)\cos(\phi) - \cos(\theta)\sin(\phi) & \cos(2\phi) \\ \sin(\theta) & \cos(\theta) & \sin(\phi) \\ -\cos(\theta) & \sin(\theta) & \cos(\phi) \end{vmatrix} \]

### Step 3: Apply Row Operations
We can perform row operations to simplify the determinant. Let's add a multiple of the second and third rows to the first row:

1. Replace \( R_1 \) with \( R_1 + \sin(\phi) R_2 + \cos(\phi) R_3 \).

After performing this operation, the first row will simplify significantly, as many terms will cancel out.

### Step 4: Expand the Determinant
Now, we can expand the determinant along the first row. The determinant can be expressed as:

\[ D = \cos(2\phi) \cdot \text{(minor of } R_1) + \text{(other terms)} \]

### Step 5: Final Simplification
After performing the necessary calculations and simplifications, we find that:

\[ D = 1 + \cos(2\phi) \]

### Conclusion
Thus, the final result for the determinant is:

\[ D = 1 + \cos(2\phi) \]