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

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

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

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

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

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

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

= \(- \sqrt{3}\)