If `sec^(2)theta-sectheta=1` , then what is the value of `(tan^(12)theta-3tan^(10)theta+3tan^(8)theta-tan^(6)theta)`?

Options

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

Answer

To solve the equation \( \sec^2 \theta - \sec \theta = 1 \) and find the value of \( \tan^{12} \theta - 3\tan^{10} \theta + 3\tan^8 \theta - \tan^6 \theta \), we can follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ \sec^2 \theta - \sec \theta - 1 = 0 \]
This is a quadratic equation in terms of \( \sec \theta \).

### Step 2: Let \( x = \sec \theta \)
Substituting \( x \) for \( \sec \theta \), we rewrite the equation as:
\[ x^2 - x - 1 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -1 \):
\[ x = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
Thus, we have:
\[ \sec \theta = \frac{1 + \sqrt{5}}{2} \quad \text{or} \quad \sec \theta = \frac{1 - \sqrt{5}}{2} \]
Since \( \sec \theta \) must be positive, we take:
\[ \sec \theta = \frac{1 + \sqrt{5}}{2} \]

### Step 4: Find \( \tan^2 \theta \)
Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \):
\[ \tan^2 \theta = \sec^2 \theta - 1 = \frac{1 + \sqrt{5}}{2}^2 - 1 \]
Calculating \( \sec^2 \theta \):
\[ \sec^2 \theta = \frac{(1 + \sqrt{5})^2}{4} = \frac{1 + 2\sqrt{5} + 5}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} \]
Thus,
\[ \tan^2 \theta = \frac{3 + \sqrt{5}}{2} - 1 = \frac{3 + \sqrt{5} - 2}{2} = \frac{1 + \sqrt{5}}{2} \]

### Step 5: Substitute \( \tan^2 \theta \) into the expression
Let \( y = \tan^2 \theta \). Then, we need to evaluate:
\[ y^6 - 3y^5 + 3y^4 - y^3 \]
Using \( y = \frac{1 + \sqrt{5}}{2} \), we can compute \( y^3, y^4, y^5, y^6 \).

### Step 6: Calculate \( y^3, y^4, y^5, y^6 \)
1. \( y^3 = y \cdot y^2 = y \cdot \frac{1 + \sqrt{5}}{2} \)
2. \( y^4 = y^2 \cdot y^2 \)
3. \( y^5 = y^4 \cdot y \)
4. \( y^6 = y^5 \cdot y \)

### Step 7: Substitute back into the polynomial
After calculating \( y^3, y^4, y^5, y^6 \), substitute these values back into the polynomial expression and simplify.

### Final Result
After performing the calculations, we find that:
\[ \tan^{12} \theta - 3\tan^{10} \theta + 3\tan^8 \theta - \tan^6 \theta = 1 \]