$7 \cdot 13$ If $\theta$ is an acute angle and $\sin \frac{\theta}{2}=\sqrt{\frac{x-1}{2 x}}$, then $\operatorname{tg} \theta$ equals
(A) $x$.
(B) $\frac{1}{x}$.
(C) $\frac{\sqrt{x-1}}{x+1}$.
(D) $\frac{\sqrt{x^{2}-1}}{x}$.
(E) $\sqrt{x^{2}-1}$.
(24th American High School Mathematics Examination, 1973)
[Solution] From $\cos \theta=1-2 \sin ^{2} \frac{\theta}{2}=1-2\left(\frac{x-1}{2 x}\right)=\frac{1}{x}$, and $\sec \theta=\frac{1}{\cos \theta}=x$,

then $\operatorname{tg}^{2} \theta=\sec ^{2} \theta-1=x^{2}-1$, so $\operatorname{tg} \theta=\sqrt{x^{2}-1}$. Therefore, the answer is $(E)$.