यदि 3 tan θ = 2 है, तो \(\frac{2 \sin\theta-\cos\theta}{2 \cos\theta-\sin\theta}\) का मान ज्ञात करें।

Question

Question

This question was previously asked in

RRB ALP CBT I 29 Aug 2018 Shift 2 Official Paper

Answer 1

दिया गया है:

3tanθ = 2

गणना:

3tanθ = 2

⇒ tan θ = 2/3

⇒ sinθ/cosθ = 2/3

⇒ sinθ = 2cosθ/3

अब,

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

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

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

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

⇒ \(\frac{1}{4}\)

∴ अभीष्ट उत्तर \(\frac{1}{4}\) है।