If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)

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 \(\frac{2}{\sqrt{3}-1}\)

waec 1999 mathematics

The correct answer is C. \(\frac{\sqrt{3}-1}{2}\)

\(\cos x = \frac{\sqrt{3}}{2}\)

\(\sin x = \frac{1}{2}\)

\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)

Previous question Next question