JAMB - Mathematics (1985 - No. 34)
If cos \(\theta\) = \(\frac{\sqrt{3}}{2}\) and \(\theta\) is less than 90o. Calculate \(\frac{\cot(90 - \theta)}{sin^2\theta}\)
\(\frac{4}{\sqrt{3}}\)
\(4 \sqrt{3}\)
\(\sqrt{\frac{3}{2}}\)
\(\frac{1}{\sqrt{3}}\)
\(\frac{2}{\sqrt{3}}\)
Explanation
\(\frac{\cot (90 - \theta)}{sin^2\theta}\)
\(\cot (90 - \theta) = \tan \theta\)
\(\therefore \frac{\cot (90 - \theta)}{\sin^{2} \theta} = \frac{\tan \theta}{\sin^{2} \theta}\)
\(\tan \theta = \frac{\sqrt{3}}{3}\)
\(\sin \theta = \frac{1}{2} \implies \sin^{2} \theta = \frac{1}{4}\)
\(\frac{\cot(90 - \theta)}{\sin^{2} \theta} = \frac{\sqrt{3}}{3}\div\frac{1}{4}\)
= \(\frac{4}{\sqrt{3}}\)
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