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.
\(\frac{2}{\sqrt{3}-1}\)
Correct answer is C
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)