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

Options

A. `sqrt((x-1)/(x+1))`
B. `(sqrt(x-1))/(x+1)`
C. `sqrt( x^(2) - 1)`
D. `sqrt( x^(2) +1)`

Answer

To solve the problem, we need to find the value of \( \tan \theta \) given that

\[ \frac{\sin \theta}{2} = \sqrt{\frac{x-1}{2x}}. \]

### Step-by-Step Solution:

1. **Square Both Sides:**
\[ \left(\frac{\sin \theta}{2}\right)^2 = \left(\sqrt{\frac{x-1}{2x}}\right)^2 \]
This simplifies to:
\[ \frac{\sin^2 \theta}{4} = \frac{x-1}{2x}. \]

2. **Multiply Both Sides by 4:**
\[ \sin^2 \theta = \frac{4(x-1)}{2x} = \frac{2(x-1)}{x}. \]

3. **Use the Pythagorean Identity:**
Recall that \( \sin^2 \theta + \cos^2 \theta = 1 \). Thus,
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{2(x-1)}{x}. \]

4. **Simplify \( \cos^2 \theta \):**
\[ \cos^2 \theta = 1 - \frac{2x - 2}{x} = \frac{x - (2x - 2)}{x} = \frac{2 - x}{x}. \]

5. **Find \( \tan \theta \):**
Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). We need to express \( \sin \theta \) and \( \cos \theta \):
- From \( \sin^2 \theta = \frac{2(x-1)}{x} \), we have \( \sin \theta = \sqrt{\frac{2(x-1)}{x}} \).
- From \( \cos^2 \theta = \frac{2-x}{x} \), we have \( \cos \theta = \sqrt{\frac{2-x}{x}} \).

6. **Substitute into \( \tan \theta \):**
\[ \tan \theta = \frac{\sqrt{\frac{2(x-1)}{x}}}{\sqrt{\frac{2-x}{x}}} = \sqrt{\frac{2(x-1)}{2-x}}. \]

7. **Final Result:**
Thus, we find that:
\[ \tan \theta = \sqrt{\frac{2(x-1)}{2-x}}. \]