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. \(\frac{2}{\sqrt{3}}\)
Correct Answer: B
Explanation
\(\frac{\cot (90 - \theta)}{sin^2\theta}\)
\(\cot (90 - \theta) = \tan \theta\)
\(\therefore \frac{\cot (90 - \theta)}{\sin^{2} \theta} = \frac{\tan \theta}{\sin^{2} \theta}\)
\(\tan \theta = \sqrt{3}\)
\(\sin \theta = \frac{1}{2} \implies \sin^{2} \theta = \frac{1}{4}\)
\(\frac{\cot(90 - \theta)}{\sin^{2} \theta} = \frac{\sqrt{3}}{\frac{1}{4}}\)
= \(4 \sqrt{3}\)
