If cos \(\theta\) = \(\frac{\sqrt{3}}{2}\) and \(\theta\) is less than 90o. Calculate \(\frac{\cot(90 - \theta)}{sin^2\theta}\)

A. \(\frac{4}{\sqrt{3}}\)
B. \(4 \sqrt{3}\)
C. \(\sqrt{\frac{3}{2}}\)
D. \(\frac{1}{\sqrt{3}}\)
E. \(\frac{2}{\sqrt{3}}\)

\(\frac{\cot (90 - \theta)}{sin^2\theta}\)

\(\cot (90 - \theta) = \tan \theta\)

\(\therefore \frac{\cot (90 - \theta)}{\sin^{2} \theta} = \frac{\tan \theta}{\sin^{2} \theta}\)

\(\tan \theta = \frac{\sqrt{3}}{3}\)

\(\sin \theta = \frac{1}{2} \implies \sin^{2} \theta = \frac{1}{4}\)

\(\frac{\cot(90 - \theta)}{\sin^{2} \theta} = \frac{\sqrt{3}}{3}\div\frac{1}{4}\)

= \(\frac{4}{\sqrt{3}}\)

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