In a `Delta ABC`, cos (A + B) + cos C =

Options

A. 2 cos C
B. `-1`
C. 0
D. 1

Answer

To solve the problem, we need to find the value of \( \cos(A + B) + \cos C \) in triangle \( ABC \).

### Step-by-Step Solution:

1. **Understand the Angles in Triangle**:
In any triangle, the sum of the angles is \( 180^\circ \). Therefore, we can write:
\[ A + B + C = 180^\circ \]

2. **Express \( A + B \)**:
From the equation above, we can express \( A + B \) as:
\[ A + B = 180^\circ - C \]

3. **Substitute into the Cosine Function**:
We want to find \( \cos(A + B) + \cos C \). Substitute \( A + B \) into the cosine function:
\[ \cos(A + B) = \cos(180^\circ - C) \]

4. **Use the Cosine Identity**:
We know from trigonometric identities that:
\[ \cos(180^\circ - \theta) = -\cos(\theta) \]
Applying this identity, we get:
\[ \cos(180^\circ - C) = -\cos C \]

5. **Combine the Results**:
Now substitute this back into our expression:
\[ \cos(A + B) + \cos C = -\cos C + \cos C \]

6. **Simplify**:
The two \( \cos C \) terms cancel each other out:
\[ -\cos C + \cos C = 0 \]

### Final Answer:
Thus, the value of \( \cos(A + B) + \cos C \) is:
\[ \boxed{0} \]