If `tan theta + sec theta = n,` then `sec ^(4) theta - tan ^(4) theta- 2 sec theta tan theta =`

Options

A. `n-1`
B. `n ^(2)`
C. `(1)/(n ^(2))`
D. `0`

Answer

To solve the problem, we need to find the value of the expression \( \sec^4 \theta - \tan^4 \theta - 2 \sec \theta \tan \theta \) given that \( \tan \theta + \sec \theta = n \).

### Step-by-Step Solution:

1. **Start with the given expression**:
\[ \sec^4 \theta - \tan^4 \theta - 2 \sec \theta \tan \theta \]

2. **Use the identity for the difference of squares**:
We can express \( \sec^4 \theta - \tan^4 \theta \) as:
\[ \sec^4 \theta - \tan^4 \theta = (\sec^2 \theta + \tan^2 \theta)(\sec^2 \theta - \tan^2 \theta) \]

3. **Recall the Pythagorean identity**:
We know that:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
Therefore, substituting this into our expression gives:
\[ \sec^4 \theta - \tan^4 \theta = (\sec^2 \theta + \tan^2 \theta)(1) = \sec^2 \theta + \tan^2 \theta \]

4. **Substituting back into the expression**:
Now we can rewrite our original expression:
\[ \sec^2 \theta + \tan^2 \theta - 2 \sec \theta \tan \theta \]

5. **Recognize another identity**:
We can express \( \sec^2 \theta + \tan^2 \theta \) as:
\[ \sec^2 \theta + \tan^2 \theta = (\sec \theta + \tan \theta)^2 - 2 \sec \theta \tan \theta \]
Thus, our expression becomes:
\[ (\sec \theta + \tan \theta)^2 - 2 \sec \theta \tan \theta - 2 \sec \theta \tan \theta \]
Simplifying this gives:
\[ (\sec \theta + \tan \theta)^2 - 4 \sec \theta \tan \theta \]

6. **Substituting \( n \)**:
Since \( \sec \theta + \tan \theta = n \), we have:
\[ n^2 - 4 \sec \theta \tan \theta \]

7. **Finding \( \sec \theta \tan \theta \)**:
We know from the identity \( \sec^2 \theta - \tan^2 \theta = 1 \) that:
\[ \sec^2 \theta + \tan^2 \theta = n^2 - 2 \tan \theta \sec \theta \]
Hence, we can find \( \sec \theta \tan \theta \) using:
\[ \sec \theta \tan \theta = \frac{n^2 - 1}{2} \]

8. **Final substitution**:
Substitute \( \sec \theta \tan \theta \) back into the expression:
\[ n^2 - 4 \left(\frac{n^2 - 1}{2}\right) = n^2 - 2(n^2 - 1) = n^2 - 2n^2 + 2 = 2 - n^2 \]

### Final Answer:
\[ \sec^4 \theta - \tan^4 \theta - 2 \sec \theta \tan \theta = 2 - n^2 \]