If `theta` is an acute angle and `sin((theta)/(2))=sqrt((x-1)/(2x))` ,then `tan theta` is equal to

Explanation

To solve the problem, we need to find the value of \( \tan \theta \) given that \( \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{x-1}{2x}} \).

### Step 1: Find \( \cos\left(\frac{\theta}{2}\right) \)
Using the Pythagorean identity, we know that:
\[ \cos^2\left(\frac{\theta}{2}\right) = 1 - \sin^2\left(\frac{\theta}{2}\right) \]
Substituting the value of \( \sin\left(\frac{\theta}{2}\right) \):
\[ \cos^2\left(\frac{\theta}{2}\right) = 1 - \left(\sqrt{\frac{x-1}{2x}}\right)^2 \]
\[ = 1 - \frac{x-1}{2x} = \frac{2x - (x - 1)}{2x} = \frac{x + 1}{2x} \]
Thus,
\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{x + 1}{2x}} \]

### Step 2: Find \( \tan\left(\frac{\theta}{2}\right) \)
Now we can find \( \tan\left(\frac{\theta}{2}\right) \):
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} = \frac{\sqrt{\frac{x-1}{2x}}}{\sqrt{\frac{x + 1}{2x}}} \]
The \( \sqrt{2x} \) cancels out:
\[ = \sqrt{\frac{x-1}{x+1}} \]

### Step 3: Find \( \tan \theta \)
Using the double angle formula for tangent:
\[ \tan \theta = \frac{2 \tan\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)} \]
Substituting \( \tan\left(\frac{\theta}{2}\right) \):
\[ \tan \theta = \frac{2 \sqrt{\frac{x-1}{x+1}}}{1 - \left(\sqrt{\frac{x-1}{x+1}}\right)^2} \]
Calculating \( \tan^2\left(\frac{\theta}{2}\right) \):
\[ \tan^2\left(\frac{\theta}{2}\right) = \frac{x-1}{x+1} \]
Thus,
\[ 1 - \tan^2\left(\frac{\theta}{2}\right) = 1 - \frac{x-1}{x+1} = \frac{(x+1) - (x-1)}{x+1} = \frac{2}{x+1} \]
Now substituting back:
\[ \tan \theta = \frac{2 \sqrt{\frac{x-1}{x+1}}}{\frac{2}{x+1}} = \sqrt{\frac{x-1}{x+1}} \cdot (x + 1) = \frac{2\sqrt{(x-1)(x+1)}}{2} = \sqrt{(x-1)(x+1)} \]

### Final Answer
Thus, the value of \( \tan \theta \) is:
\[ \tan \theta = \sqrt{(x-1)(x+1)} \]