Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o – 1}{1 – tan 30^o}\)
- \(\sqrt{3 - 2}\)
- 2 - \(\sqrt{3}\)
-
\(\sqrt{3}\)
- -2
The correct answer is: C
Explanation
From the figure above, tan30º = \(\frac{1}{\sqrt3}\) and tan60º = \(\sqrt{3}\)
so \(\frac{tan60 - 1}{1 - tan30}\)
= \(\frac{\sqrt{3}}{1}\div(1 - (\frac{1}{\sqrt{3}}\)))
= \(\frac{\sqrt{3} - 1}{1}\div\frac{\sqrt{3} - 1}{\sqrt3}\)
= \(\frac{\sqrt{3} - 1}{1}\times \frac{\sqrt3}{\sqrt{3} - 1}\) = \(\sqrt{3}\)