If 3 cot θ = 2, find the value of \(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\).
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If 3 cot θ = 2, find the value of \(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\).
If 3 cot θ = 2, find the value of \(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\).
Given,
3 cot θ = 2
⇒ cot θ = \(\frac{2}{3}\)
\(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\)
From, let’s divide the numerator and denominator by sin θ. We get,
\(\frac{(4 –3 cot θ) }{(2 + 6 cot θ)}\)
⇒ \(\frac{(4 – 3(2/3)) }{(2 + 6(2/3))}\) [using the value of tan θ]
⇒ \(\frac{(4 – 2)}{(2 + 4)}\) [After taking LCM and simplifying it]
⇒\(\frac{2}{6}\)
= \(\frac{1}{3}\)
\(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\) = \(\frac{1}{3}\).