If `tan^(2)theta, sec^(2)theta` are the roots of `ax^(2)+bx+c=0` then `b^(2)- a^(2) =`

Options

A. 4ac
B. `a^(2)`
C. `4bc`
D. 4ab

Answer

To solve the problem, we need to find the value of \( b^2 - a^2 \) given that \( \tan^2 \theta \) and \( \sec^2 \theta \) are the roots of the quadratic equation \( ax^2 + bx + c = 0 \).

### Step-by-step Solution:

1. **Identify the Roots**:
Let \( \alpha = \tan^2 \theta \) and \( \beta = \sec^2 \theta \).

2. **Use the Sum and Product of Roots**:
From the properties of quadratic equations, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

3. **Calculate the Sum of Roots**:
\[ \alpha + \beta = \tan^2 \theta + \sec^2 \theta \]
Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \), we can rewrite:
\[ \tan^2 \theta + \sec^2 \theta = \tan^2 \theta + (1 + \tan^2 \theta) = 2\tan^2 \theta + 1 \]
Thus, we have:
\[ -\frac{b}{a} = 2\tan^2 \theta + 1 \]

4. **Calculate the Product of Roots**:
\[ \alpha \beta = \tan^2 \theta \cdot \sec^2 \theta = \tan^2 \theta \cdot (1 + \tan^2 \theta) = \tan^2 \theta + \tan^4 \theta \]
Thus, we have:
\[ \frac{c}{a} = \tan^2 \theta + \tan^4 \theta \]

5. **Use the Identity**:
We can use the identity:
\[ (\alpha + \beta)^2 = (\alpha - \beta)^2 + 4\alpha\beta \]
Substituting the values:
\[ \left(-\frac{b}{a}\right)^2 = (\sec^2 \theta - \tan^2 \theta)^2 + 4\left(\frac{c}{a}\right) \]

6. **Evaluate \( \sec^2 \theta - \tan^2 \theta \)**:
Using the identity \( \sec^2 \theta - \tan^2 \theta = 1 \):
\[ \left(-\frac{b}{a}\right)^2 = 1 + 4\left(\frac{c}{a}\right) \]
Thus:
\[ \frac{b^2}{a^2} = 1 + \frac{4c}{a} \]

7. **Cross-Multiply**:
\[ b^2 = a^2 + 4ac \]

8. **Rearranging**:
Now, we can rearrange this to find \( b^2 - a^2 \):
\[ b^2 - a^2 = 4ac \]

### Final Result:
Thus, the value of \( b^2 - a^2 \) is:
\[ \boxed{4ac} \]