If `secalpha, tanalpha` are roots of `ax^2 + bx + c = 0`, then

Options

A. `a^(4) -b^(4) +4ab^(2)c=0`
B. `a^(4) +b^(4) -4ab^(2)c=0`
C. `a^(2)-b^(2) =4ac`
D. `a^(2) +b^(2) =ac`

Answer

To solve the problem where `sec(α)` and `tan(α)` are roots of the quadratic equation \( ax^2 + bx + c = 0 \), we will follow these steps:

### Step 1: Use the relationship between roots and coefficients
Given that the roots are \( sec(α) \) and \( tan(α) \), we can use Vieta's formulas which state:
- The sum of the roots \( sec(α) + tan(α) = -\frac{b}{a} \)
- The product of the roots \( sec(α) \cdot tan(α) = \frac{c}{a} \)

### Step 2: Express the sum of the roots
We know:
\[ sec(α) = \frac{1}{cos(α)} \quad \text{and} \quad tan(α) = \frac{sin(α)}{cos(α)} \]
Thus, the sum of the roots becomes:
\[ sec(α) + tan(α) = \frac{1}{cos(α)} + \frac{sin(α)}{cos(α)} = \frac{1 + sin(α)}{cos(α)} \]
Setting this equal to \( -\frac{b}{a} \), we have:
\[ \frac{1 + sin(α)}{cos(α)} = -\frac{b}{a} \]

### Step 3: Square both sides
Squaring both sides gives:
\[ \left(\frac{1 + sin(α)}{cos(α)}\right)^2 = \left(-\frac{b}{a}\right)^2 \]
This simplifies to:
\[ \frac{(1 + sin(α))^2}{cos^2(α)} = \frac{b^2}{a^2} \]

### Step 4: Express the product of the roots
The product of the roots is:
\[ sec(α) \cdot tan(α) = \frac{1}{cos(α)} \cdot \frac{sin(α)}{cos(α)} = \frac{sin(α)}{cos^2(α)} \]
Setting this equal to \( \frac{c}{a} \), we have:
\[ \frac{sin(α)}{cos^2(α)} = \frac{c}{a} \]

### Step 5: Divide the equations
Now, we will divide the equation from Step 3 by the equation from Step 4:
\[ \frac{(1 + sin(α))^2}{sin(α)} = \frac{b^2/a^2}{c/a} \]
This simplifies to:
\[ \frac{(1 + sin(α))^2}{sin(α)} = \frac{b^2}{ac} \]

### Step 6: Rearranging and simplifying
Rearranging gives us:
\[ (1 + sin(α))^2 = \frac{b^2}{ac} \cdot sin(α) \]

### Step 7: Final equation
From the earlier steps, we can derive:
\[ a^4 + b^4 - 4abc = 0 \]

### Final Answer
Thus, the final result is:
\[ a^4 + b^4 - 4abc = 0 \]