The value of `F(x)=6cosxsqrt(1+tan^2x) + 2sinxsqrt(1+cot^2x)` where `x in (0,2pi)-{pi,pi/2,(3pi)/2}` may be

Options

A. 4
B. `-4`
C. 8
D. `-8`

Answer

To solve the problem, we need to evaluate the function:

\[ F(x) = 6 \cos x \sqrt{1 + \tan^2 x} + 2 \sin x \sqrt{1 + \cot^2 x} \]

We can simplify this using trigonometric identities.

### Step 1: Simplify the square root terms

Using the identities:
- \( 1 + \tan^2 x = \sec^2 x \)
- \( 1 + \cot^2 x = \csc^2 x \)

We can rewrite the function as:

\[ F(x) = 6 \cos x \sqrt{\sec^2 x} + 2 \sin x \sqrt{\csc^2 x} \]

### Step 2: Substitute the square roots

Since \( \sqrt{\sec^2 x} = |\sec x| \) and \( \sqrt{\csc^2 x} = |\csc x| \), we can write:

\[ F(x) = 6 \cos x |\sec x| + 2 \sin x |\csc x| \]

### Step 3: Rewrite using trigonometric functions

Now, since \( |\sec x| = \frac{1}{|\cos x|} \) and \( |\csc x| = \frac{1}{|\sin x|} \), we have:

\[ F(x) = 6 \cos x \cdot \frac{1}{|\cos x|} + 2 \sin x \cdot \frac{1}{|\sin x|} \]

This simplifies to:

\[ F(x) = 6 \cdot \text{sgn}(\cos x) + 2 \cdot \text{sgn}(\sin x) \]

where \( \text{sgn}(y) \) is the sign function which returns -1 for negative values, 0 for zero, and +1 for positive values.

### Step 4: Analyze the function over the intervals

Now we need to analyze \( F(x) \) over the intervals \( (0, 2\pi) \) excluding \( \{\pi, \frac{\pi}{2}, \frac{3\pi}{2}\} \):

1. **Interval \( (0, \frac{\pi}{2}) \)**:
- \( \cos x > 0 \) and \( \sin x > 0 \)
- \( F(x) = 6(1) + 2(1) = 8 \)

2. **Interval \( (\frac{\pi}{2}, \pi) \)**:
- \( \cos x < 0 \) and \( \sin x > 0 \)
- \( F(x) = 6(-1) + 2(1) = -6 + 2 = -4 \)

3. **Interval \( (pi, \frac{3\pi}{2}) \)**:
- \( \cos x < 0 \) and \( \sin x < 0 \)
- \( F(x) = 6(-1) + 2(-1) = -6 - 2 = -8 \)

4. **Interval \( (\frac{3\pi}{2}, 2\pi) \)**:
- \( \cos x > 0 \) and \( \sin x < 0 \)
- \( F(x) = 6(1) + 2(-1) = 6 - 2 = 4 \)

### Step 5: Conclusion

The possible values of \( F(x) \) over the specified intervals are:

- \( 8 \) for \( (0, \frac{\pi}{2}) \)
- \( -4 \) for \( (\frac{\pi}{2}, \pi) \)
- \( -8 \) for \( (pi, \frac{3\pi}{2}) \)
- \( 4 \) for \( (\frac{3\pi}{2}, 2\pi) \)

Thus, the final answer is that \( F(x) \) may take values \( 8, -4, -8, 4 \).