`sec^(6)theta-tan^(6)theta-3tan^(2)theta.sec^(2)theta=?`

Options

A. 1
B. 3
C. 2
D. `-1`

Answer

To solve the expression \( \sec^6 \theta - \tan^6 \theta - 3 \tan^2 \theta \sec^2 \theta \), we can use the identity for the difference of cubes.

### Step-by-Step Solution:

1. **Recognize the Identity**:
We can use the identity for the difference of cubes:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Here, let \( a = \sec^2 \theta \) and \( b = \tan^2 \theta \). Thus, we can rewrite the expression as:
\[ \sec^6 \theta - \tan^6 \theta = (\sec^2 \theta - \tan^2 \theta)(\sec^4 \theta + \sec^2 \theta \tan^2 \theta + \tan^4 \theta) \]

2. **Simplify \( \sec^2 \theta - \tan^2 \theta \)**:
We know from trigonometric identities that:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
Therefore, we can substitute this into our expression:
\[ \sec^6 \theta - \tan^6 \theta = 1 \cdot (\sec^4 \theta + \sec^2 \theta \tan^2 \theta + \tan^4 \theta) = \sec^4 \theta + \sec^2 \theta \tan^2 \theta + \tan^4 \theta \]

3. **Substituting Back into the Original Expression**:
Now we substitute this back into the original expression:
\[ \sec^4 \theta + \sec^2 \theta \tan^2 \theta + \tan^4 \theta - 3 \tan^2 \theta \sec^2 \theta \]

4. **Combine Like Terms**:
Combine the terms involving \( \tan^2 \theta \sec^2 \theta \):
\[ \sec^4 \theta + (\sec^2 \theta \tan^2 \theta - 3 \tan^2 \theta \sec^2 \theta) + \tan^4 \theta \]
This simplifies to:
\[ \sec^4 \theta - 2 \tan^2 \theta \sec^2 \theta + \tan^4 \theta \]

5. **Factor the Expression**:
Notice that the expression can be factored:
\[ \sec^4 \theta - 2 \tan^2 \theta \sec^2 \theta + \tan^4 \theta = (\sec^2 \theta - \tan^2 \theta)^2 \]
Since we already established that \( \sec^2 \theta - \tan^2 \theta = 1 \):
\[ (\sec^2 \theta - \tan^2 \theta)^2 = 1^2 = 1 \]

### Final Answer:
Thus, the final result is:
\[ \sec^6 \theta - \tan^6 \theta - 3 \tan^2 \theta \sec^2 \theta = 1 \]