If ` x= npi+(-1)^(n)alpha , n in I and x=npi+(-1)^(n)beta ` are the roots of ` 4 cos x -3 sec x =tan x `, then `4(sin alpha + sin beta )` is

Options

A. `-1`
B. 1
C. 2
D. None of these

Answer

To solve the given problem, we need to find the value of \( 4(\sin \alpha + \sin \beta) \) where \( x = n\pi + (-1)^n \alpha \) and \( x = n\pi + (-1)^n \beta \) are the roots of the equation \( 4 \cos x - 3 \sec x = \tan x \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ 4 \cos x - 3 \sec x = \tan x \]

2. **Rewrite the equation using trigonometric identities**:
Recall that \( \sec x = \frac{1}{\cos x} \) and \( \tan x = \frac{\sin x}{\cos x} \). Substitute these into the equation:
\[ 4 \cos x - \frac{3}{\cos x} = \frac{\sin x}{\cos x} \]

3. **Multiply through by \( \cos x \) to eliminate the fractions** (assuming \( \cos x \neq 0 \)):
\[ 4 \cos^2 x - 3 = \sin x \]

4. **Use the identity \( \sin^2 x + \cos^2 x = 1 \)** to express \( \cos^2 x \) in terms of \( \sin x \):
\[ \cos^2 x = 1 - \sin^2 x \]
Substitute this into the equation:
\[ 4(1 - \sin^2 x) - 3 = \sin x \]

5. **Simplify the equation**:
\[ 4 - 4 \sin^2 x - 3 = \sin x \]
\[ 1 - 4 \sin^2 x - \sin x = 0 \]

6. **Rearrange the equation**:
\[ 4 \sin^2 x + \sin x - 1 = 0 \]

7. **Use the quadratic formula to solve for \( \sin x \)**:
The quadratic formula is given by:
\[ \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = 1 \), and \( c = -1 \):
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
\[ \sin x = \frac{-1 \pm \sqrt{1 + 16}}{8} \]
\[ \sin x = \frac{-1 \pm \sqrt{17}}{8} \]

8. **Identify the values of \( \alpha \) and \( \beta \)**:
From the roots, we have:
\[ \alpha = \sin^{-1}\left(\frac{-1 + \sqrt{17}}{8}\right), \quad \beta = \sin^{-1}\left(\frac{-1 - \sqrt{17}}{8}\right) \]

9. **Calculate \( 4(\sin \alpha + \sin \beta) \)**:
Since \( \sin \alpha = \frac{-1 + \sqrt{17}}{8} \) and \( \sin \beta = \frac{-1 - \sqrt{17}}{8} \):
\[ \sin \alpha + \sin \beta = \frac{-1 + \sqrt{17}}{8} + \frac{-1 - \sqrt{17}}{8} \]
\[ = \frac{-2}{8} = -\frac{1}{4} \]

10. **Final calculation**:
\[ 4(\sin \alpha + \sin \beta) = 4 \left(-\frac{1}{4}\right) = -1 \]

### Conclusion:
The value of \( 4(\sin \alpha + \sin \beta) \) is \( -1 \).