Question on: WAEC Mathematics - 1999
If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)
A
\(\frac{\sqrt{3}+1}{2}\)
B
\(\frac{2}{\sqrt{3}+1}\)
C
\(\frac{\sqrt{3}-1}{2}\)
D
\(\frac{2}{\sqrt{3}-1}\)
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Correct Option: C
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)
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