If `x epsilon(7pi,8pi)`, then `"tan"^(-1)sqrt((1-"cos" x)/(1+"cos" x))=`

Options

A. `-x/2`
B. `x/2`
C. `4pi-x/2`
D. None of these

Answer

To solve the problem, we need to evaluate the expression \( \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}} \) for \( x \in (7\pi, 8\pi) \).

### Step-by-Step Solution:

1. **Identify the Range for \( x \)**:
Given \( x \in (7\pi, 8\pi) \), we can find the range for \( \frac{x}{2} \):
\[ \frac{7\pi}{2} < \frac{x}{2} < \frac{8\pi}{2} \implies \frac{7\pi}{2} < \frac{x}{2} < 4\pi \]
This gives us \( \frac{x}{2} \in \left(\frac{7\pi}{2}, 4\pi\right) \).

**Hint**: Divide the range of \( x \) by 2 to find the corresponding range for \( \frac{x}{2} \).

2. **Use Trigonometric Identities**:
We can simplify the expression \( \sqrt{\frac{1 - \cos x}{1 + \cos x}} \) using the half-angle identities:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \quad \text{and} \quad 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Substituting these into the expression gives:
\[ \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \sqrt{\frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)}} = \sqrt{\tan^2\left(\frac{x}{2}\right)} = |\tan\left(\frac{x}{2}\right)| \]

**Hint**: Use half-angle identities to simplify trigonometric expressions.

3. **Determine the Sign of \( \tan\left(\frac{x}{2}\right) \)**:
Since \( \frac{x}{2} \) lies in the range \( \left(\frac{7\pi}{2}, 4\pi\right) \), we can determine the quadrant:
- \( \frac{7\pi}{2} = 3.5\pi \) is in the third quadrant.
- \( 4\pi \) is at the boundary of the fourth quadrant.
Therefore, \( \frac{x}{2} \) is in the fourth quadrant where \( \tan\left(\frac{x}{2}\right) < 0 \).

**Hint**: Determine the quadrant based on the angle to find the sign of the tangent function.

4. **Evaluate the Inverse Tangent**:
Since \( \tan\left(\frac{x}{2}\right) < 0 \), we have:
\[ \sqrt{\frac{1 - \cos x}{1 + \cos x}} = -\tan\left(\frac{x}{2}\right) \]
Thus, we can write:
\[ \tan^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) = \tan^{-1}\left(-\tan\left(\frac{x}{2}\right)\right) \]

5. **Use the Property of Inverse Tangent**:
Using the property \( \tan^{-1}(-y) = -\tan^{-1}(y) \):
\[ \tan^{-1}\left(-\tan\left(\frac{x}{2}\right)\right) = -\tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) = -\frac{x}{2} \]

**Hint**: Use properties of inverse functions to simplify the expression.

6. **Final Expression**:
Therefore, the final result is:
\[ \tan^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) = -\frac{x}{2} \]

### Conclusion:
The value of \( \tan^{-1}\sqrt{\frac{1 - \cos x}{1 + \cos x}} \) for \( x \in (7\pi, 8\pi) \) is \( -\frac{x}{2} \).