If \(tan x = \frac{1}{\sqrt{3}}\), find cos x – sin x such that \(0^o \leq x \leq 90^o\)
- \(\frac{\sqrt{3}+1}{2}\)
- \(\frac{2}{\sqrt{3}+1}\)
-
\(\frac{\sqrt{3}-1}{2}\)
- \(\frac{2}{\sqrt{3}-1}\)
The correct answer is: C
Explanation
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)