smc

Since $\frac{\theta }{2}$ is acute and $\cos \frac{\theta }{2}=\sqrt{1-\sin (\frac{\theta }{2}{)}^2}$, $$\cos \frac{\theta }{2}=\sqrt{1-(\frac{x-1}{2x}{)}^2}$$ $$\cos \frac{\theta }{2}=\sqrt{\frac{x+1}{2x}}$$ Using the definition of tangent, $$\tan \frac{\theta }{2}=\sqrt{\frac{x-1}{2x}}÷\sqrt{\frac{x+1}{2x}}$$ $$\tan \frac{\theta }{2}=\sqrt{\frac{x-1}{x+1}}$$ 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 $\boxed{\sqrt{x^2-1}}$.