Question on: WAEC Mathematics - 2009
Given that sin 60o = (\frac{\sqrt{3}}{2}) and cos 60o = (\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 Option: D
Sin 60 = (\frac{\sqrt{3}}{2}); cos 60o = (\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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