The determinant `D=|{:(cos(alpha+beta),-sin(alpha+beta),cos2beta),(sinalpha,cosalpha,sinbeta),(-cosalpha,sinalpha,cosbeta):}|` is independent of :-

Options

A. `alpha`
B. `beta`
C. `alpha" and "beta`
D. Neihter `alpha"nor"beta`

Answer

To solve the given determinant \( D = \begin{vmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) & \cos(2\beta) \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta \end{vmatrix} \), we will follow a systematic approach to simplify it and determine its independence from the variables involved.

### Step 1: Write the determinant
We start with the determinant:
\[ D = \begin{vmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) & \cos(2\beta) \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta \end{vmatrix} \]

### Step 2: Expand the determinant
We can expand the determinant using the first row:
\[ D = \cos(\alpha + \beta) \begin{vmatrix} \cos \alpha & \sin \beta \\ \sin \alpha & \cos \beta \end{vmatrix} - (-\sin(\alpha + \beta)) \begin{vmatrix} \sin \alpha & \sin \beta \\ -\cos \alpha & \cos \beta \end{vmatrix} + \cos(2\beta) \begin{vmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} \cos \alpha & \sin \beta \\ \sin \alpha & \cos \beta \end{vmatrix} = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta) \)
2. \( \begin{vmatrix} \sin \alpha & \sin \beta \\ -\cos \alpha & \cos \beta \end{vmatrix} = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \sin(\alpha + \beta) \)
3. \( \begin{vmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{vmatrix} = \sin^2 \alpha + \cos^2 \alpha = 1 \)

### Step 4: Substitute back into the determinant
Substituting these results back into the expression for \( D \):
\[ D = \cos(\alpha + \beta) \cdot \cos(\alpha + \beta) + \sin(\alpha + \beta) \cdot \sin(\alpha + \beta) + \cos(2\beta) \cdot 1 \]
\[ D = \cos^2(\alpha + \beta) + \sin^2(\alpha + \beta) + \cos(2\beta) \]

### Step 5: Apply Pythagorean identity
Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \):
\[ D = 1 + \cos(2\beta) \]

### Step 6: Simplify further
Using the double angle formula \( \cos(2\beta) = 2\cos^2(\beta) - 1 \):
\[ D = 1 + (2\cos^2(\beta) - 1) = 2\cos^2(\beta) \]

### Conclusion
The final expression for the determinant is:
\[ D = 2\cos^2(\beta) \]
This shows that \( D \) is independent of \( \alpha \) and depends only on \( \beta \).