# 1973 AHSME Problems/Problem 17

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## Problem

If $\theta$ is an acute angle and $\sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}$, then $\tan \theta$ equals

$\textbf{(A)}\ x \qquad \textbf{(B)}\ \frac1{x} \qquad \textbf{(C)}\ \frac{\sqrt{x-1}}{x+1} \qquad \textbf{(D)}\ \frac{\sqrt{x^2-1}}{x} \qquad \textbf{(E)}\ \sqrt{x^2-1}$

## Solution

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 $\boxed{\textbf{(E)}}$.