If `sec^(4)theta-sec^(2)theta=3` then the value of `tan^(4)theta+tan^(2)theta` is:

Options

A. 8
B. 4
C. 6
D. 3

Answer

To solve the equation \( \sec^4 \theta - \sec^2 \theta = 3 \) and find the value of \( \tan^4 \theta + \tan^2 \theta \), we can follow these steps:

### Step 1: Substitute \( z = \sec^2 \theta \)
Let \( z = \sec^2 \theta \). Then the equation becomes:
\[ z^4 - z^2 = 3 \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ z^4 - z^2 - 3 = 0 \]

### Step 3: Let \( w = z^2 \)
Now, let \( w = z^2 \). The equation can be rewritten as:
\[ w^2 - w - 3 = 0 \]

### Step 4: Solve the quadratic equation
We can use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -1, c = -3 \):
\[ w = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]
\[ w = \frac{1 \pm \sqrt{1 + 12}}{2} \]
\[ w = \frac{1 \pm \sqrt{13}}{2} \]

### Step 5: Find \( z^2 \)
Since \( w = z^2 \), we have:
\[ z^2 = \frac{1 + \sqrt{13}}{2} \quad \text{or} \quad z^2 = \frac{1 - \sqrt{13}}{2} \]
Since \( z^2 \) must be positive, we take:
\[ z^2 = \frac{1 + \sqrt{13}}{2} \]

### Step 6: Find \( \tan^2 \theta \)
Using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \):
\[ \tan^2 \theta = z^2 - 1 = \frac{1 + \sqrt{13}}{2} - 1 = \frac{-1 + \sqrt{13}}{2} \]

### Step 7: Calculate \( \tan^4 \theta + \tan^2 \theta \)
Now, we need to find \( \tan^4 \theta + \tan^2 \theta \):
\[ \tan^4 \theta = (\tan^2 \theta)^2 = \left(\frac{-1 + \sqrt{13}}{2}\right)^2 \]
Calculating \( \tan^4 \theta \):
\[ \tan^4 \theta = \frac{(-1 + \sqrt{13})^2}{4} = \frac{1 - 2\sqrt{13} + 13}{4} = \frac{14 - 2\sqrt{13}}{4} = \frac{7 - \sqrt{13}}{2} \]

Now, add \( \tan^2 \theta \):
\[ \tan^4 \theta + \tan^2 \theta = \frac{7 - \sqrt{13}}{2} + \frac{-1 + \sqrt{13}}{2} \]
Combining the fractions:
\[ = \frac{(7 - \sqrt{13}) + (-1 + \sqrt{13})}{2} = \frac{6}{2} = 3 \]

### Final Answer
Thus, the value of \( \tan^4 \theta + \tan^2 \theta \) is:
\[ \boxed{3} \]