If 3 tan θ = 2, find the value of \(\frac{2 \sin\theta-\cos\theta}{2 \cos\theta-\sin\theta}\).

Question

Question

This question was previously asked in

RRB ALP CBT I 29 Aug 2018 Shift 2 Official Paper

Answer 1

Given:

3tanθ = 2

Calculation:

3tanθ = 2

⇒ tanθ = 2/3

⇒ sinθ/cosθ = 2/3

⇒ sinθ = 2cosθ/3

Now,

\(\frac{2 \sin\theta-\cos\theta}{2 \cos\theta-\sin\theta}\) = \(\frac{2\times\frac{2cos\theta}{3}-\cos\theta}{2 \cos\theta-\frac{2cos\theta}{3}}\)

⇒ \(\frac{\frac{4cos\theta}{3}-\cos\theta}{2 \cos\theta-\frac{2cos\theta}{3}}\)

⇒ \(\frac{\frac{4cos\theta - 3cos\theta}{3}}{\frac{6cos\theta - 2cos\theta}{3}}\)

⇒ \(\frac{\frac{cos\theta}{3}}{\frac{4cos\theta}{3}}\)

⇒ \(\frac{1}{4}\)

∴ The required answer is \(\frac{1}{4}\).

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