In a `DeltaABC,angleC=90^(@)`, then the equation whose roots are tan A & tan B is

Options

A. `abx^(2)+c^(2)x+ab=0`
B. `abx^(2)+c^(2)x-ab=0`
C. `abx^(2)-c^(2)x-ab=0`
D. `abx^(2)-c^(2)x+ab=0`

Answer

To solve the problem, we need to find the quadratic equation whose roots are \( \tan A \) and \( \tan B \) in a right triangle \( \Delta ABC \) where \( \angle C = 90^\circ \).

### Step-by-Step Solution:

1. **Identify the Triangle and Angles**:
In triangle \( ABC \), since \( \angle C = 90^\circ \), we can denote the sides opposite to angles \( A \) and \( B \) as follows:
- Let \( a \) be the length of side \( BC \) (opposite angle \( A \)).
- Let \( b \) be the length of side \( AC \) (opposite angle \( B \)).
- Let \( c \) be the length of side \( AB \) (the hypotenuse).

2. **Express \( \tan A \) and \( \tan B \)**:
Using the definitions of tangent in a right triangle:
- \( \tan A = \frac{a}{b} \)
- \( \tan B = \frac{b}{a} \)

3. **Identify the Roots**:
Let \( \alpha = \tan A = \frac{a}{b} \) and \( \beta = \tan B = \frac{b}{a} \).

4. **Calculate \( \alpha + \beta \)**:
\[ \alpha + \beta = \frac{a}{b} + \frac{b}{a} \]
To combine these fractions, find a common denominator:
\[ \alpha + \beta = \frac{a^2 + b^2}{ab} \]

5. **Use Pythagorean Theorem**:
In a right triangle, we know:
\[ a^2 + b^2 = c^2 \]
Therefore:
\[ \alpha + \beta = \frac{c^2}{ab} \]

6. **Calculate \( \alpha \cdot \beta \)**:
\[ \alpha \cdot \beta = \tan A \cdot \tan B = \frac{a}{b} \cdot \frac{b}{a} = 1 \]

7. **Form the Quadratic Equation**:
The quadratic equation with roots \( \alpha \) and \( \beta \) can be expressed as:
\[ x^2 - (\alpha + \beta)x + (\alpha \cdot \beta) = 0 \]
Substituting the values we found:
\[ x^2 - \left(\frac{c^2}{ab}\right)x + 1 = 0 \]

8. **Multiply by \( ab \)**:
To eliminate the fraction, multiply the entire equation by \( ab \):
\[ abx^2 - c^2x + ab = 0 \]

### Final Quadratic Equation:
The quadratic equation whose roots are \( \tan A \) and \( \tan B \) is:
\[ abx^2 - c^2x + ab = 0 \]