WAEC Further Mathematics 2011 Paper

Concept:Use the exact value of $\sin 60^\circ$ and rationalize the denominator.Explanation:Recall that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.Substitute this into the denominator of the given expression.The expression becomes $\frac{1}{1 - \frac{\sqrt{3}}{2}}$.Combine the terms in the denominator: $1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$.Thus, the fraction becomes $\frac{1}{\frac{2 - \sqrt{3}}{2}} = \frac{2}{2 - \sqrt{3}}$.Rationalize the denominator by multiplying by the conjugate $2 + \sqrt{3}$.So, $\frac{2}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}$.The denominator simplifies to $4 - 3 = 1$.Hence, the value is $\frac{2(2 + \sqrt{3})}{1} = 4 + 2\sqrt{3}$.Answer:$4 + 2\sqrt{3}$, which is option D.