If `x + y =z`, find the value of `cos^2 x + cos^2 y +cos^2z`- 2cosx.cosy.cosz.

Options

A. a.cos^2z
B. b.sin^2z
C. c.0
D. d.1

Answer

To solve the problem, we need to find the value of \( \cos^2 x + \cos^2 y + \cos^2 z - 2 \cos x \cos y \cos z \) given that \( x + y = z \).

### Step-by-Step Solution:

1. **Substitute \( z \)**:
Since \( z = x + y \), we can rewrite \( \cos z \) as \( \cos(x + y) \).
\[ \cos z = \cos(x + y) = \cos x \cos y - \sin x \sin y \]

2. **Express \( \cos^2 z \)**:
Now we can find \( \cos^2 z \):
\[ \cos^2 z = (\cos x \cos y - \sin x \sin y)^2 \]
Expanding this using the square of a binomial:
\[ \cos^2 z = \cos^2 x \cos^2 y - 2 \cos x \cos y \sin x \sin y + \sin^2 x \sin^2 y \]

3. **Substituting into the original expression**:
Now we substitute \( \cos^2 z \) back into the expression we want to evaluate:
\[ \cos^2 x + \cos^2 y + \cos^2 z - 2 \cos x \cos y \cos z \]
becomes:
\[ \cos^2 x + \cos^2 y + \left(\cos^2 x \cos^2 y - 2 \cos x \cos y \sin x \sin y + \sin^2 x \sin^2 y\right) - 2 \cos x \cos y (\cos x \cos y - \sin x \sin y) \]

4. **Simplifying the expression**:
Now we simplify the expression:
\[ = \cos^2 x + \cos^2 y + \cos^2 x \cos^2 y - 2 \cos x \cos y \sin x \sin y + \sin^2 x \sin^2 y - 2 \cos^2 x \cos^2 y + 2 \cos x \cos y \sin x \sin y \]
Notice that the \( -2 \cos x \cos y \sin x \sin y \) and \( +2 \cos x \cos y \sin x \sin y \) cancel out:
\[ = \cos^2 x + \cos^2 y - \cos^2 x \cos^2 y + \sin^2 x \sin^2 y \]

5. **Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \)**:
We can use the identity \( \sin^2 x = 1 - \cos^2 x \) and \( \sin^2 y = 1 - \cos^2 y \):
\[ = \cos^2 x + \cos^2 y - \cos^2 x \cos^2 y + (1 - \cos^2 x)(1 - \cos^2 y) \]
Expanding the last term:
\[ = \cos^2 x + \cos^2 y - \cos^2 x \cos^2 y + 1 - \cos^2 x - \cos^2 y + \cos^2 x \cos^2 y \]

6. **Final simplification**:
All terms involving \( \cos^2 x \) and \( \cos^2 y \) cancel out:
\[ = 1 \]

### Conclusion:
The final value is:
\[ \boxed{1} \]