JAMB 1985 · UME · Q38

If $\cos\theta = \frac{\sqrt3}{2}$ and $\theta$ is less than $90^\circ$ , calculate $\dfrac{\cot(90^\circ - \theta)}{\sin^2\theta}$ .

Worked solution (try it first)

$\cos\theta = \frac{\sqrt3}{2}$ with $\theta$ acute, so $\theta = 30^\circ$ .
Then $\cot(90^\circ - \theta) = \cot60^\circ$
$= \frac{1}{\sqrt3}$ , and $\sin^2 30^\circ = \left(\frac12\right)^2$
$= \frac14$ .
Divide: $\dfrac{1/\sqrt3}{1/4} = \dfrac{4}{\sqrt3}$ .
Rationalise by multiplying top and bottom by $\sqrt3$ : $\dfrac{4\sqrt3}{3}$ , option A.