If $\theta$ is an acute angle and $\sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}$, then $\tan \theta$ equals 
$\textbf{(A)}\ x \qquad \textbf{(B)}\ \frac1{x} \qquad \textbf{(C)}\ \frac{\sqrt{x-1}}{x+1} \qquad \textbf{(D)}\ \frac{\sqrt{x^2-1}}{x} \qquad \textbf{(E)}\ \sqrt{x^2-1}$
Since $\tfrac{\theta}{2}$ is acute and $\cos \tfrac{\theta}{2} = \sqrt{1 - \sin (\tfrac{\theta}{2})^2}$,
\[\cos \frac{\theta}{2} = \sqrt{1 - (\frac{x-1}{2x})^2}\]
\[\cos \frac{\theta}{2} = \sqrt{\frac{x+1}{2x}}\]
Using the definition of tangent,
\[\tan \frac{\theta}{2} = \sqrt{\frac{x-1}{2x}} \div \sqrt{\frac{x+1}{2x}}\]
\[\tan \frac{\theta}{2} = \sqrt{\frac{x-1}{x+1}}\]
Finally, by using the Double Angle Identity for Tangent,
\[\tan \theta = \frac{2 \cdot \sqrt{\frac{x-1}{x+1}}}{1 - \frac{x-1}{x+1}}\]
\[\tan \theta = \frac{\frac{2\sqrt{x^2-1}}{x+1}}{\frac{2}{x+1}}\]
\[\tan \theta = \sqrt{x^2-1}\]
The answer is $\boxed{\textbf{(E)}}$.