If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq...
MATHEMATICS
WAEC 1999
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: C. \(\frac{\sqrt{3}-1}{2}\)
Explanation
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)
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