If θ is an acute angle and sin12θ=√(x-1)/(2x), then tanθ equals

1973 AMC 12 Problem 17

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17.

\theta

is an acute angle and

\sin\frac12\theta=\sqrt{\frac{x-1}{2x}},

then

\tan\theta

equals

x

\dfrac1x

\dfrac{\sqrt{x-1}}{x+1}

\dfrac{\sqrt{x^2-1}}x

\sqrt{x^2-1}

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.