Question : If $2\cot x=5$, then what is $\frac{2 \cos x-\sin x}{2 \cos x+\sin x}$ equal to?
Option 1: $\frac{3}{4}$
Option 2: $\frac{1}{3}$
Option 3: $\frac{5}{6}$
Option 4: $\frac{2}{3}$
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Correct Answer: $\frac{2}{3}$
Solution :
$2\cot x=5$
Or, $\cot x = \frac{5}{2}$
Given expression:
$\frac{2 \cos x-\sin x}{2 \cos x+\sin x}$
Dividing the numerator and the denominator by $\sin x $,
The expression becomes,
$\frac{2\cot x - 1}{2\cot x + 1} = \frac{2\times\frac{5}{2} - 1}{2\times\frac{5}{2}+1}=\frac{5-1}{5+1}=\frac{4}{6} = \frac{2}{3}$
Hence, the correct answer is $\frac{2}{3}$.
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