waecmaths question:
Given that$\cos {{30}^{\circ }}=\sin {{60}^{\circ }}=\frac{\sqrt{3}}{2}$ and $\sin {{30}^{\circ }}=\cos {{60}^{\circ }}=\frac{1}{2}$ evaluate $\frac{\tan {{60}^{\circ }}-1}{1-\tan {{30}^{\circ }}}$
Option A:
$\sqrt{3}-2$
Option B:
$2-\sqrt{3}$
Option C:
$\sqrt{3}$
Option D:
–2
waecmaths solution:
$\begin{align} & \frac{\tan {{60}^{\circ }}-1}{1-\tan {{30}^{\circ }}}=\frac{\tfrac{\sin {{60}^{\circ }}}{\cos {{60}^{\circ }}}-1}{1-\tfrac{\sin {{30}^{\circ }}}{\cos {{30}^{\circ }}}}=\frac{\tfrac{\tfrac{\sqrt{3}}{2}}{\tfrac{1}{2}}-1}{1-\tfrac{\tfrac{1}{2}}{\tfrac{\sqrt{3}}{2}}}=\frac{\sqrt{3}-1}{1-\tfrac{1}{\sqrt{3}}} \\ & \frac{\tan {{60}^{\circ }}-1}{1-\tan {{30}^{\circ }}}=\frac{\sqrt{3}-1}{\tfrac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{1}\times \frac{\sqrt{3}}{\sqrt{3}-1}=\sqrt{3} \\\end{align}$
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