WAEC - Mathematics (1999 - No. 19)
If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)
\(\frac{\sqrt{3}+1}{2}\)
\(\frac{2}{\sqrt{3}+1}\)
\(\frac{\sqrt{3}-1}{2}\)
\(\frac{2}{\sqrt{3}-1}\)
ব্যাখ্যা
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)