Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o - q}{1 - tan 30^o}\)

  • A \(\sqrt{3 - 2}\)
  • B 2 - \(\sqrt{3}\)
  • C \(\sqrt{3}\)
  • D -2
waec 2017mathematics

The correct answer is C. \(\sqrt{3}\)

Tan 60 = 3; Tan 30 = 1

\(\frac{\tan 60^o - 1}{1 - tan 30^o}\)  = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\)

= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\)

= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\sqrt{3}\)

Previous question Next question