Given that sin 60o = \(\frac{\sqrt{3}}{2}\) and cos 60o = \(\frac{1}{2}\), evaluate \(\frac{1 - sin...

WAEC 2009

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

A. \(\frac{2 + \sqrt{3}}{3}\)
B. \(\frac{1 - \sqrt{3}}{3}\)
C. \(\frac{1 + \sqrt{3}}{3}\)
D. \(\frac{2 - \sqrt{3}}{3}\)

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Correct Answer: Option D

Explanation

Sin 60 = \(\frac{\sqrt{3}}{2}\); cos 60^o = \(\frac{1}{2}\)

= \(\frac{1 - \sin 60}{1 + \cos 60} = \frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{1}{2}}\)

= \(\frac{2 - \sqrt{3}}{3}{2}\div \frac{2 + 1}{2}\)

= \(\frac{2 - \sqrt{3}}{2} \times \frac{2}{3}\)

= \(\frac{2 - \sqrt{3}}{3}\)

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