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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