If `tan A, tanB` are the roots of the quadratic `abx^2-c^2x+ab=0`, where a,b,c are the sides of a triangle , then

Options

A. `tan A = a/b`
B. `tan B =b/a`
C. `cos C=0`
D. `tan A+ tanB=c^2/(ab)`

Answer

To solve the problem, we need to analyze the quadratic equation given and the relationships between the roots and the sides of the triangle.

### Step-by-Step Solution:

1. **Identify the Quadratic Equation**:
The given quadratic equation is \( abx^2 - c^2x + ab = 0 \).

2. **Roots of the Quadratic**:
Let the roots of the quadratic be \( \tan A \) and \( \tan B \). According to Vieta's formulas, the sum and product of the roots can be expressed in terms of the coefficients of the quadratic equation.

3. **Sum of Roots**:
From Vieta's formulas, the sum of the roots \( \tan A + \tan B \) is given by:
\[ \tan A + \tan B = \frac{c^2}{ab} \]

4. **Product of Roots**:
The product of the roots \( \tan A \tan B \) is given by:
\[ \tan A \tan B = \frac{ab}{ab} = 1 \]

5. **Using the Tangent Addition Formula**:
We know that:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting the values we found:
\[ \tan(A + B) = \frac{\frac{c^2}{ab}}{1 - 1} = \text{undefined} \]
This implies that \( A + B = 90^\circ \) (or \( \frac{\pi}{2} \) radians).

6. **Using the Cosine Rule**:
Since \( A + B + C = 180^\circ \), we have:
\[ C = 90^\circ \]
This indicates that the triangle is a right triangle.

7. **Finding \( \cos C \)**:
For a right triangle, \( \cos C = 0 \) since \( C = 90^\circ \).

8. **Conclusion**:
The relationships derived from the roots of the quadratic indicate that \( C \) is the right angle of the triangle formed by sides \( a, b, c \).

### Final Result:
The triangle is a right triangle with \( C = 90^\circ \).