1973 AMC 12 第 17 题

Use $\cos\theta=1-2\sin^2(\frac{\theta}{2})$

Once $\cos\theta$ is known, apply $\tan^2\theta=\sec^2\theta-1$ and use that $\theta$ is acute

The half-angle identity gives $$\begin{aligned} \cos\theta &=1-2\sin^2\frac{\theta}{2}\\ &=1-\frac{x-1}{x}\\ &=\frac1x. \end{aligned}$$ Therefore $$\tan^2\theta=\sec^2\theta-1=x^2-1.$$ Because $\theta$ is acute, its tangent is positive, so $$\tan\theta=\sqrt{x^2-1}.$$

Therefore, the correct answer is E.