In `sinA and cosA` are the roots of the equation `4x^(2)-3x+a=0, sinA+cosA+tanA+cotA+secA+"cosec "A=7 and 0 lt A lt (pi)/(2)`, then the value of a must be

Options

A. `(7)/(25)`
B. `(25)/(7)`
C. `(28)/(25)`
D. `(25)/(28)`

Answer

To solve the problem, we need to find the value of \( a \) given that \( \sin A \) and \( \cos A \) are the roots of the equation \( 4x^2 - 3x + a = 0 \) and that \( \sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7 \).

### Step 1: Use Vieta's Formulas
Since \( \sin A \) and \( \cos A \) are the roots of the quadratic equation \( 4x^2 - 3x + a = 0 \), we can apply Vieta's formulas.

- The sum of the roots \( \sin A + \cos A = -\frac{B}{A} = \frac{3}{4} \).
- The product of the roots \( \sin A \cdot \cos A = \frac{C}{A} = \frac{a}{4} \).

### Step 2: Express \( \tan A \), \( \cot A \), \( \sec A \), and \( \csc A \)
We know the following relationships:
- \( \tan A = \frac{\sin A}{\cos A} \)
- \( \cot A = \frac{\cos A}{\sin A} \)
- \( \sec A = \frac{1}{\cos A} \)
- \( \csc A = \frac{1}{\sin A} \)

### Step 3: Substitute these into the equation
The equation becomes:
\[ \sin A + \cos A + \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} + \frac{1}{\cos A} + \frac{1}{\sin A} = 7 \]
Let \( S = \sin A + \cos A \) and \( P = \sin A \cdot \cos A \).

### Step 4: Rewrite the equation
Using \( S = \frac{3}{4} \) and \( P = \frac{a}{4} \):
\[ S + \frac{S^2}{P} + \frac{1}{\cos A} + \frac{1}{\sin A} = 7 \]
We can express \( \frac{1}{\cos A} + \frac{1}{\sin A} \) as:
\[ \frac{1}{\cos A} + \frac{1}{\sin A} = \frac{\sin A + \cos A}{\sin A \cos A} = \frac{S}{P} \]
Thus, the equation simplifies to:
\[ S + \frac{S^2}{P} + \frac{S}{P} = 7 \]

### Step 5: Substitute \( S \) and \( P \)
Substituting \( S = \frac{3}{4} \):
\[ \frac{3}{4} + \frac{\left(\frac{3}{4}\right)^2}{\frac{a}{4}} + \frac{\frac{3}{4}}{\frac{a}{4}} = 7 \]
This simplifies to:
\[ \frac{3}{4} + \frac{\frac{9}{16}}{\frac{a}{4}} + \frac{\frac{3}{4}}{\frac{a}{4}} = 7 \]
Multiplying through by \( 4a \) to eliminate the fractions:
\[ 3a + \frac{9}{4} + 3 = 28a \]
This leads to:
\[ 3a + \frac{21}{4} = 28a \]
Rearranging gives:
\[ \frac{21}{4} = 28a - 3a \]
\[ \frac{21}{4} = 25a \]

### Step 6: Solve for \( a \)
Thus,
\[ a = \frac{21}{4 \cdot 25} = \frac{21}{100} \]

### Final Answer
The value of \( a \) must be \( \frac{21}{100} \).