If 3 cot θ = 2, find the value of \(\frac{4sin\, θ\, -\, 3cos\, θ}

1.	If 3 cot θ = 2, find the value of \(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\).
Answer» 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] ⇒\(\frac{2}{6}\) = \(\frac{1}{3}\) \(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\) = \(\frac{1}{3}\).

If 3 cot θ = 2, find the value of \(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\).

Answer»

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]

⇒\(\frac{2}{6}\)

= \(\frac{1}{3}\)

\(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\) = \(\frac{1}{3}\).