`tan ^(-1)(sqrt((1-cos 3 x)/(1+cos x)))`

Answer

To solve the expression \( \tan^{-1} \left( \sqrt{\frac{1 - \cos 3x}{1 + \cos 3x}} \right) \), we can follow these steps:

### Step 1: Use Trigonometric Identities
We start with the expression inside the inverse tangent function:
\[ \sqrt{\frac{1 - \cos 3x}{1 + \cos 3x}} \]
We can use the trigonometric identities:
- \( 1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right) \)
- \( 1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right) \)

### Step 2: Apply the Identities
Using these identities for \( \theta = 3x \):
\[ 1 - \cos 3x = 2 \sin^2 \left(\frac{3x}{2}\right) \]
\[ 1 + \cos 3x = 2 \cos^2 \left(\frac{3x}{2}\right) \]
Substituting these into our expression gives:
\[ \sqrt{\frac{2 \sin^2 \left(\frac{3x}{2}\right)}{2 \cos^2 \left(\frac{3x}{2}\right)}} = \sqrt{\frac{\sin^2 \left(\frac{3x}{2}\right)}{\cos^2 \left(\frac{3x}{2}\right)}} \]

### Step 3: Simplify the Expression
This simplifies to:
\[ \sqrt{\tan^2 \left(\frac{3x}{2}\right)} = |\tan \left(\frac{3x}{2}\right)| \]
Since the range of \( \tan^{-1} \) is from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), we can drop the absolute value:
\[ \tan^{-1} \left( |\tan \left(\frac{3x}{2}\right)| \right) = \tan^{-1} \left( \tan \left(\frac{3x}{2}\right) \right) \]

### Step 4: Final Result
Thus, we have:
\[ \tan^{-1} \left( \tan \left(\frac{3x}{2}\right) \right) = \frac{3x}{2} \]
Therefore, the final answer is:
\[ \frac{3x}{2} \]