Given that sin 60o frac sqrt 3 2 and cos 60o fr... - WAEC Mathematics 2009 Question
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
\(\frac{2 - \sqrt{3}}{3}\)
correct option: d
Sin 60 = \(\frac{\sqrt{3}}{2}\); cos 60o = \(\frac{1}{2}\)
= \(\frac{1 - \sin 60}{1 + \cos 60} = \frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{1}{2}}\)
= \(\frac{2 - \sqrt{3}}{3}{2}\div \frac{2 + 1}{2}\)
= \(\frac{2 - \sqrt{3}}{2} \times \frac{2}{3}\)
= \(\frac{2 - \sqrt{3}}{3}\)
= \(\frac{1 - \sin 60}{1 + \cos 60} = \frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{1}{2}}\)
= \(\frac{2 - \sqrt{3}}{3}{2}\div \frac{2 + 1}{2}\)
= \(\frac{2 - \sqrt{3}}{2} \times \frac{2}{3}\)
= \(\frac{2 - \sqrt{3}}{3}\)
Please share this, thanks:
Add your answer
No responses