If `angle A and angle B` are complementary to each other,then the value of `sec^2 A + sec^2 B - sec^2 A sec^2 B` is

Options

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

Answer

To solve the problem, we need to find the value of \( \sec^2 A + \sec^2 B - \sec^2 A \sec^2 B \) given that angles \( A \) and \( B \) are complementary. This means \( A + B = 90^\circ \) or \( B = 90^\circ - A \).

### Step-by-Step Solution:

1. **Express \( B \) in terms of \( A \)**:
Since \( A \) and \( B \) are complementary, we can write:
\[ B = 90^\circ - A \]

2. **Substitute \( B \) into the expression**:
We need to substitute \( B \) into the expression \( \sec^2 A + \sec^2 B - \sec^2 A \sec^2 B \):
\[ \sec^2 A + \sec^2(90^\circ - A) - \sec^2 A \sec^2(90^\circ - A) \]

3. **Use the identity for \( \sec(90^\circ - A) \)**:
We know that:
\[ \sec(90^\circ - A) = \csc A \]
Therefore, we have:
\[ \sec^2(90^\circ - A) = \csc^2 A \]

4. **Rewrite the expression**:
Now we can rewrite our expression as:
\[ \sec^2 A + \csc^2 A - \sec^2 A \csc^2 A \]

5. **Use the identities for \( \sec^2 A \) and \( \csc^2 A \)**:
Recall the identities:
\[ \sec^2 A = 1 + \tan^2 A \quad \text{and} \quad \csc^2 A = 1 + \cot^2 A \]

6. **Substitute these identities**:
Substitute these into the expression:
\[ (1 + \tan^2 A) + (1 + \cot^2 A) - (1 + \tan^2 A)(1 + \cot^2 A) \]

7. **Simplify the expression**:
Expanding the product:
\[ (1 + \tan^2 A)(1 + \cot^2 A) = 1 + \tan^2 A + \cot^2 A + \tan^2 A \cot^2 A \]
Thus, our expression becomes:
\[ 2 + \tan^2 A + \cot^2 A - (1 + \tan^2 A + \cot^2 A + \tan^2 A \cot^2 A) \]

8. **Combine like terms**:
This simplifies to:
\[ 2 - 1 - \tan^2 A \cot^2 A = 1 - \tan^2 A \cot^2 A \]

9. **Recognize that \( \tan^2 A \cot^2 A = 1 \)**:
Since \( \tan A \cdot \cot A = 1 \), we have:
\[ \tan^2 A \cot^2 A = 1 \]

10. **Final simplification**:
Therefore, we can conclude:
\[ 1 - 1 = 0 \]

### Conclusion:
The final value of \( \sec^2 A + \sec^2 B - \sec^2 A \sec^2 B \) is \( 0 \).