Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o - q}{1 - tan 30^o}\)
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Correct Answer: Option C
Explanation:
Tan 60 = 3; Tan 30 = 1
\(\frac{\tan 60^o - 1}{1 - tan 30^o}\) = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= \(\sqrt{3}\)
Tan 60 = 3; Tan 30 = 1
\(\frac{\tan 60^o - 1}{1 - tan 30^o}\) = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= \(\sqrt{3}\)