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
\(\frac{2}{\sqrt{3}-1}\)
Verified Correct Answer: C.
Step-by-Step Mathematical Solution
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)
Source: Myschool Past Questions Archive (Record #26399)