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

Options

A. 1
B. -1
C. 0
D. none of these

Answer

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

### Step-by-Step Solution:

1. **Understanding the Properties of a Cyclic Quadrilateral**:
- In a cyclic quadrilateral, the sum of the opposite angles is \( 180^\circ \). Therefore, we have:
\[ A + C = 180^\circ \quad \text{and} \quad B + D = 180^\circ \]

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

3. **Substituting C and D into the Expression**:
- Now, we substitute these values into the expression \( \cos A + \cos B + \cos C + \cos D \):
\[ \cos A + \cos B + \cos(180^\circ - A) + \cos(180^\circ - B) \]

4. **Using the Cosine Identity**:
- We use the identity \( \cos(180^\circ - \theta) = -\cos \theta \):
\[ \cos(180^\circ - A) = -\cos A \quad \text{and} \quad \cos(180^\circ - B) = -\cos B \]

5. **Simplifying the Expression**:
- Substituting these identities back into our expression gives:
\[ \cos A + \cos B - \cos A - \cos B \]
- This simplifies to:
\[ 0 \]

6. **Conclusion**:
- Therefore, the value of \( \cos A + \cos B + \cos C + \cos D \) is:
\[ \boxed{0} \]