Suppose ABCS (in order) is a quadrilateral inscribed in a circle. Which of the following is/are always true? `secB=secD` (b) `cotA+cotC=0` `cos e cA=cos e cC` (d) `tanB+tanD=0`

Options

A. `secB=secB`
B. `cotA+cotC=0`
C. `cosecA=cosecC`
D. `tanB+tanD=0`

Explanation

To solve the problem, we need to analyze the properties of a cyclic quadrilateral (a quadrilateral inscribed in a circle) and check the validity of each given statement.

### Step-by-Step Solution:

1. **Understanding the Properties of Cyclic Quadrilaterals:**
- In a cyclic quadrilateral, the sum of opposite angles is supplementary:
- \( A + C = 180^\circ \)
- \( B + D = 180^\circ \)

2. **Analyzing Option (a): \( \sec B = \sec D \)**
- We know that \( D = 180^\circ - B \).
- Using the secant function:
\[ \sec D = \sec(180^\circ - B) = -\sec B \]
- Thus, \( \sec B \neq \sec D \).
- **Conclusion:** Option (a) is **incorrect**.

3. **Analyzing Option (b): \( \cot A + \cot C = 0 \)**
- Since \( C = 180^\circ - A \):
- We can write:
\[ \cot C = \cot(180^\circ - A) = -\cot A \]
- Therefore:
\[ \cot A + \cot C = \cot A - \cot A = 0 \]
- **Conclusion:** Option (b) is **correct**.

4. **Analyzing Option (c): \( \csc A = \csc C \)**
- Again, since \( C = 180^\circ - A \):
- We can write:
\[ \csc C = \csc(180^\circ - A) = \csc A \]
- Therefore:
\[ \csc A = \csc C \]
- **Conclusion:** Option (c) is **correct**.

5. **Analyzing Option (d): \( \tan B + \tan D = 0 \)**
- Since \( D = 180^\circ - B \):
- We can write:
\[ \tan D = \tan(180^\circ - B) = -\tan B \]
- Therefore:
\[ \tan B + \tan D = \tan B - \tan B = 0 \]
- **Conclusion:** Option (d) is **correct**.

### Final Answers:
The correct options are:
- (b) \( \cot A + \cot C = 0 \)
- (c) \( \csc A = \csc C \)
- (d) \( \tan B + \tan D = 0 \)