In a cyclic quadrilateral ABCD,
then find `cosA+cosB+cosC+cosD=?`

Answer

To solve the problem of finding the value of \( \cos A + \cos B + \cos C + \cos D \) in a cyclic quadrilateral ABCD, we can follow these steps:

### Step-by-Step Solution:

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

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

3. **Substituting into the Cosine Sum**:
Now, we substitute these expressions into the sum \( \cos A + \cos B + \cos C + \cos D \):
\[ \cos A + \cos B + \cos C + \cos D = \cos(180^\circ - C) + \cos(180^\circ - D) + \cos C + \cos D \]

4. **Using the Cosine Identity**:
We use the identity \( \cos(180^\circ - \theta) = -\cos \theta \):
\[ \cos(180^\circ - C) = -\cos C \quad \text{and} \quad \cos(180^\circ - D) = -\cos D \]
Thus, we can rewrite the expression:
\[ = -\cos C - \cos D + \cos C + \cos D \]

5. **Simplifying the Expression**:
Now, we simplify the expression:
\[ -\cos C + \cos C - \cos D + \cos D = 0 \]

6. **Final Result**:
Therefore, we conclude that:
\[ \cos A + \cos B + \cos C + \cos D = 0 \]

### Final Answer:
\[ \cos A + \cos B + \cos C + \cos D = 0 \]