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Given \(\sin \theta =  \frac{\sqrt{3}}{2}, 0° \leq ...

Given \sin \theta =  \frac{\sqrt{3}}{2}, 0° \leq \theta \leq 90°, find \tan 2\theta in surd form

A.

- \sqrt{3}

B.

-\frac{\sqrt{3}}{2}

C.

\frac{\sqrt{3}}{2}

D.

\sqrt{3}

Correct answer is A

\sin \theta = \frac{\sqrt{3}}{2} \implies opp = \sqrt{3}; hyp = 2

adj^{2} = 2^{2} - (\sqrt{3})^{2} = 1 \implies adj = 1

\cos \theta = \frac{1}{2}

\sin 2\theta = \sin (180 - \theta) = \sin \theta = \frac{\sqrt{3}}{2}

\cos 2\theta = \cos (180 - \theta) = -\cos \theta = -\frac{1}{2}

\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}

= - \sqrt{3}