If ABCD is a convex quadrilateral such that `4 sec A+5 =0` then the quadratic equation whose roots are tan A and cosec A is

Options

A. `12x^(2)-29x+15=0`
B. `12x^(2)-11x-15=0`
C. `12x^(2)+11x-15=0`
D. none of these

Answer

To solve the problem, we need to find the quadratic equation whose roots are \( \tan A \) and \( \csc A \) given that \( 4 \sec A + 5 = 0 \).

### Step-by-Step Solution:

1. **Find \( \sec A \)**:
\[ 4 \sec A + 5 = 0 \implies 4 \sec A = -5 \implies \sec A = -\frac{5}{4} \]

2. **Find \( \cos A \)**:
Since \( \sec A = \frac{1}{\cos A} \):
\[ \cos A = -\frac{4}{5} \]

3. **Find \( \sin A \)**:
Using the identity \( \sin^2 A + \cos^2 A = 1 \):
\[ \sin^2 A = 1 - \cos^2 A = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
Thus,
\[ \sin A = \pm \frac{3}{5} \]
Since \( A \) is in the second quadrant (as \( \cos A < 0 \)), we have:
\[ \sin A = \frac{3}{5} \]

4. **Find \( \tan A \)**:
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \]

5. **Find \( \csc A \)**:
\[ \csc A = \frac{1}{\sin A} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]

6. **Form the quadratic equation**:
Let the roots be \( r_1 = \tan A = -\frac{3}{4} \) and \( r_2 = \csc A = \frac{5}{3} \).
The sum of the roots \( S \) and product of the roots \( P \) are given by:
\[ S = r_1 + r_2 = -\frac{3}{4} + \frac{5}{3} \]
To add these fractions, find a common denominator (which is 12):
\[ S = -\frac{9}{12} + \frac{20}{12} = \frac{11}{12} \]

The product of the roots:
\[ P = r_1 \cdot r_2 = \left(-\frac{3}{4}\right) \left(\frac{5}{3}\right) = -\frac{15}{12} = -\frac{5}{4} \]

7. **Write the quadratic equation**:
The standard form of a quadratic equation with roots \( r_1 \) and \( r_2 \) is:
\[ x^2 - Sx + P = 0 \]
Substituting \( S \) and \( P \):
\[ x^2 - \frac{11}{12}x - \frac{5}{4} = 0 \]
To eliminate fractions, multiply through by 12:
\[ 12x^2 - 11x - 15 = 0 \]

### Final Answer:
The quadratic equation whose roots are \( \tan A \) and \( \csc A \) is:
\[ 12x^2 - 11x - 15 = 0 \]