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

$\frac{2 + \sqrt{3}}{3}$

$\frac{1 - \sqrt{3}}{3}$

$\frac{1 + \sqrt{3}}{3}$

$\frac{2 - \sqrt{3}}{3}$

Correct answer is D

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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