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 \(0^o \leq x \leq 90^o\)

Answer Details

Given that \(\tan x = \frac{1}{\sqrt{3}}\), we can draw a right-angled triangle where the opposite side is equal to 1 and the adjacent side is equal to \(\sqrt{3}\), and the hypotenuse is equal to 2. Therefore, sin x = \(\frac{1}{2}\) and cos x = \(\frac{\sqrt{3}}{2}\). Now, cos x - sin x = \(\frac{\sqrt{3}}{2}\) - \(\frac{1}{2}\) = \(\frac{\sqrt{3}-1}{2}\). Hence, the correct option is \(\frac{\sqrt{3}-1}{2}\).