A line makes angles `alpha, beta, gamma` with the coordinate axes. If `alpha + beta =(pi)/(2), " then " (cos alpha + cos beta+ cos gamma )^(2)` is equal to

Options

A. `1 + cos 2alpha`
B. `1- sin 2alpha`
C. `1+ sin 2 alpha`
D. None of these

Answer

To solve the problem, we need to find the value of \((\cos \alpha + \cos \beta + \cos \gamma)^2\) given that \(\alpha + \beta = \frac{\pi}{2}\).

### Step-by-Step Solution:

1. **Understanding the Angles**:
Given that \(\alpha + \beta = \frac{\pi}{2}\), we can express \(\beta\) in terms of \(\alpha\):
\[ \beta = \frac{\pi}{2} - \alpha \]

2. **Using the Cosine Identity**:
We know that:
\[ \cos(\frac{\pi}{2} - \alpha) = \sin(\alpha) \]
Therefore, we can substitute \(\beta\):
\[ \cos \beta = \sin \alpha \]

3. **Using the Cosine of Gamma**:
The angles \(\alpha\), \(\beta\), and \(\gamma\) are related to the coordinate axes, and we know that:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]
Substituting \(\cos \beta\):
\[ \cos^2 \alpha + \sin^2 \alpha + \cos^2 \gamma = 1 \]

4. **Applying the Pythagorean Identity**:
From the Pythagorean identity, we know:
\[ \cos^2 \alpha + \sin^2 \alpha = 1 \]
Thus, we can simplify:
\[ 1 + \cos^2 \gamma = 1 \]
This implies:
\[ \cos^2 \gamma = 0 \]
Therefore:
\[ \cos \gamma = 0 \]

5. **Calculating the Expression**:
Now we can substitute back into our expression:
\[ L = (\cos \alpha + \sin \alpha + 0)^2 \]
This simplifies to:
\[ L = (\cos \alpha + \sin \alpha)^2 \]

6. **Expanding the Square**:
Expanding the square gives:
\[ L = \cos^2 \alpha + \sin^2 \alpha + 2 \cos \alpha \sin \alpha \]
Using the Pythagorean identity again:
\[ L = 1 + 2 \cos \alpha \sin \alpha \]

7. **Using the Double Angle Identity**:
We know that:
\[ 2 \cos \alpha \sin \alpha = \sin(2\alpha) \]
Therefore, we can rewrite \(L\) as:
\[ L = 1 + \sin(2\alpha) \]

### Final Result:
Thus, the value of \((\cos \alpha + \cos \beta + \cos \gamma)^2\) is:
\[ L = 1 + \sin(2\alpha) \]