If ABCD is a cyclic quadrilateral, then `cos A + cos B ` is equal to

Options

A. 0
B. `cos C + cos D`
C. `- (cos C+ cos D)`
D. `cos C- cos D`

Answer

To solve the problem, we need to find the value of \( \cos A + \cos B \) for a cyclic quadrilateral ABCD.

### Step-by-step Solution:

1. **Understanding Cyclic Quadrilateral Properties**:
In a cyclic quadrilateral, the opposite angles are supplementary. Thus, we have:
\[ A + C = 180^\circ \quad \text{and} \quad B + D = 180^\circ \]

2. **Expressing Angles**:
From the supplementary angle property, we can express angle \( C \) and angle \( D \) in terms of angles \( A \) and \( B \):
\[ C = 180^\circ - A \quad \text{and} \quad D = 180^\circ - B \]

3. **Using Cosine Angle Identity**:
We can use the cosine identity for angles:
\[ \cos(180^\circ - \theta) = -\cos(\theta) \]
Applying this to angles \( C \) and \( D \):
\[ \cos C = \cos(180^\circ - A) = -\cos A \]
\[ \cos D = \cos(180^\circ - B) = -\cos B \]

4. **Substituting Back into the Expression**:
Now substituting these values into the expression \( \cos A + \cos B \):
\[ \cos A + \cos B = -\cos C - \cos D \]

5. **Final Expression**:
Therefore, we can conclude that:
\[ \cos A + \cos B = -(\cos C + \cos D) \]

6. **Conclusion**:
Since \( \cos C + \cos D \) can be expressed in terms of \( \cos A + \cos B \), we can say:
\[ \cos A + \cos B = -(\cos C + \cos D) \]
This shows the relationship between the cosines of the angles in a cyclic quadrilateral.

### Final Answer:
Thus, the value of \( \cos A + \cos B \) is equal to:
\[ -\cos C - \cos D \]