Home » » If $$tan x = \frac{1}{\sqrt{3}}, find cos x - sin x$$such that $$0^o \le x \ge 90^o$$...

# If $$tan x = \frac{1}{\sqrt{3}}, find cos x - sin x$$such that $$0^o \le x \ge 90^o$$...

### Question

If $$tan x = \frac{1}{\sqrt{3}}, find cos x - sin x$$such that $$0^o \le x \ge 90^o$$

### Options

A) $$\frac{\sqrt{3}+1}{2}$$

B) $$\frac{2}{\sqrt{3}+1}$$

C) $$\frac{\sqrt{3}-1}{2}$$

D) $$\frac{2}{\sqrt{3}-1}$$