waecmaths question:
Given that $\sin {{60}^{\circ }}=\tfrac{\sqrt{3}}{2}\text{ and }\cos {{60}^{\circ }}=\tfrac{1}{2}$ evaluate $\frac{1-\sin {{60}^{\circ }}}{1+\cos {{60}^{\circ }}}$
Option A:
$\frac{2+\sqrt{3}}{3}$
Option B:
$\frac{1-\sqrt{3}}{3}$
Option C:
$\frac{1+\sqrt{3}}{3}$
Option D:
$\frac{2-\sqrt{3}}{3}$
waecmaths solution:
$\frac{1-\sin {{60}^{\circ }}}{1+\cos {{60}^{\circ }}}=\frac{1-\tfrac{\sqrt{3}}{2}}{1+\tfrac{1}{2}}=\frac{\tfrac{2-\sqrt{3}}{2}}{\tfrac{2+1}{2}}=\frac{2-\sqrt{3}}{3}$
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