If `alpha` is an acute angle and `sin(alpha/2)=sqrt((x-1)/(2x))` then `tan alpha` is

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 \( \tan \alpha \) given that \( \sin \left( \frac{\alpha}{2} \right) = \sqrt{\frac{x-1}{2x}} \).

### Step-by-Step Solution:

1. **Use the Half-Angle Identity for Cosine**:
We know that:
\[ \cos \alpha = 1 - 2 \sin^2 \left( \frac{\alpha}{2} \right) \]
Substituting the given value of \( \sin \left( \frac{\alpha}{2} \right) \):
\[ \cos \alpha = 1 - 2 \left( \sqrt{\frac{x-1}{2x}} \right)^2 \]

2. **Calculate \( \sin^2 \left( \frac{\alpha}{2} \right) \)**:
\[ \sin^2 \left( \frac{\alpha}{2} \right) = \frac{x-1}{2x} \]
Thus,
\[ \cos \alpha = 1 - 2 \cdot \frac{x-1}{2x} = 1 - \frac{x-1}{x} \]

3. **Simplify \( \cos \alpha \)**:
\[ \cos \alpha = 1 - \left(1 - \frac{1}{x}\right) = \frac{1}{x} \]

4. **Find \( \sin \alpha \)**:
Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ \sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{1}{x}\right)^2 \]
Therefore,
\[ \sin \alpha = \sqrt{1 - \frac{1}{x^2}} = \sqrt{\frac{x^2 - 1}{x^2}} = \frac{\sqrt{x^2 - 1}}{x} \]

5. **Calculate \( \tan \alpha \)**:
We know that:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]
Substituting the values we found:
\[ \tan \alpha = \frac{\frac{\sqrt{x^2 - 1}}{x}}{\frac{1}{x}} = \sqrt{x^2 - 1} \]

### Final Answer:
\[ \tan \alpha = \sqrt{x^2 - 1} \]