If `|{:(1,cos alpha, cos beta),(cos alpha, 1 , cos gamma ),(cos beta, cos gamma , 1):}|=|{:(0,cos alpha, cos beta),(cos alpha , 0 , cos gamma),(cos beta, cos gamma, 0):}|` then the value of `cos^2 alpha + cos^2 beta + cos^2 gamma` is : (a) `1` (b) `1/2` (c) `3/8` (d) `9/4`

Options

A. 1
B. `1/2`
C. `3/8`
D. `9/4`

Answer

To solve the problem, we need to evaluate the determinants given in the question and set them equal to each other.

### Step-by-Step Solution:

1. **Set up the determinants:**
We have two determinants:
\[ D_1 = \begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} \]
and
\[ D_2 = \begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{vmatrix} \]

2. **Expand \(D_1\):**
We will expand \(D_1\) along the first row:
\[ D_1 = 1 \cdot \begin{vmatrix} 1 & \cos \gamma \\ \cos \gamma & 1 \end{vmatrix} - \cos \alpha \cdot \begin{vmatrix} \cos \alpha & \cos \gamma \\ \cos \beta & 1 \end{vmatrix} + \cos \beta \cdot \begin{vmatrix} \cos \alpha & 1 \\ \cos \beta & \cos \gamma \end{vmatrix} \]

Calculating the first determinant:
\[ \begin{vmatrix} 1 & \cos \gamma \\ \cos \gamma & 1 \end{vmatrix} = 1 - \cos^2 \gamma = \sin^2 \gamma \]

Now for the second determinant:
\[ \begin{vmatrix} \cos \alpha & \cos \gamma \\ \cos \beta & 1 \end{vmatrix} = \cos \alpha \cdot 1 - \cos \beta \cdot \cos \gamma = \cos \alpha - \cos \beta \cos \gamma \]

And for the third determinant:
\[ \begin{vmatrix} \cos \alpha & 1 \\ \cos \beta & \cos \gamma \end{vmatrix} = \cos \alpha \cdot \cos \gamma - \cos \beta \cdot 1 = \cos \alpha \cos \gamma - \cos \beta \]

Putting it all together:
\[ D_1 = \sin^2 \gamma - \cos \alpha (\cos \alpha - \cos \beta \cos \gamma) + \cos \beta (\cos \alpha \cos \gamma - \cos \beta) \]

Simplifying:
\[ D_1 = \sin^2 \gamma - \cos^2 \alpha + \cos \alpha \cos \beta \cos \gamma + \cos \beta \cos \alpha \cos \gamma - \cos^2 \beta \]
\[ D_1 = \sin^2 \gamma - \cos^2 \alpha - \cos^2 \beta + 2 \cos \alpha \cos \beta \cos \gamma \]

3. **Expand \(D_2\):**
We will expand \(D_2\) along the first row:
\[ D_2 = 0 - \cos \alpha \cdot \begin{vmatrix} \cos \alpha & \cos \gamma \\ \cos \beta & 0 \end{vmatrix} + \cos \beta \cdot \begin{vmatrix} \cos \alpha & 0 \\ \cos \beta & \cos \gamma \end{vmatrix} \]

The first determinant:
\[ \begin{vmatrix} \cos \alpha & \cos \gamma \\ \cos \beta & 0 \end{vmatrix} = 0 - \cos \beta \cos \gamma = -\cos \beta \cos \gamma \]

The second determinant:
\[ \begin{vmatrix} \cos \alpha & 0 \\ \cos \beta & \cos \gamma \end{vmatrix} = \cos \alpha \cos \gamma \]

Thus:
\[ D_2 = \cos \alpha \cos \beta \cos \gamma + \cos \beta \cdot \cos \alpha \cos \gamma = 2 \cos \alpha \cos \beta \cos \gamma \]

4. **Set the determinants equal:**
We have:
\[ D_1 = D_2 \]
Therefore:
\[ \sin^2 \gamma - \cos^2 \alpha - \cos^2 \beta + 2 \cos \alpha \cos \beta \cos \gamma = 2 \cos \alpha \cos \beta \cos \gamma \]

Simplifying gives:
\[ \sin^2 \gamma - \cos^2 \alpha - \cos^2 \beta = 0 \]

5. **Using the identity:**
We know that:
\[ \sin^2 \gamma = 1 - \cos^2 \gamma \]
Substituting this gives:
\[ 1 - \cos^2 \gamma - \cos^2 \alpha - \cos^2 \beta = 0 \]
Rearranging gives:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

Thus, the value of \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma \) is \( \boxed{1} \).