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

1.	If 3 tan θ = 4, find the value of \(\frac{4 cos\, θ\, -\, sin\, θ}{2 cos\, θ\, +\, sin\, θ}\).
Answer» Given, 3 tan θ = 4 ⇒ tan θ = \(\frac{4}{3}\) \(\frac{4 cos\, θ\, -\, sin\, θ}{2 cos\, θ\, +\, sin\, θ}\) From, let’s divide the numerator and denominator by cos θ. We get, \(\frac{(4\, –\, tan\, θ)}{(2\, +\, tan\, θ)}\) ⇒ \(\frac{(4 – (4/3))}{(2 + (4/3))}\) [using the value of tan θ] ⇒ \(\frac{(12 – 4)}{(6 + 4)}\) [After taking LCM and cancelling it] ⇒ \(\frac{8}{10}\) = \(\frac{4}{5}\) \(\frac{4 cos\, θ\, -\, sin\, θ}{2 cos\, θ\, +\, sin\, θ}\) = \(\frac{4}{5}\)

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

Answer»

Given,

3 tan θ = 4

⇒ tan θ = \(\frac{4}{3}\)

\(\frac{4 cos\, θ\, -\, sin\, θ}{2 cos\, θ\, +\, sin\, θ}\)

From, let’s divide the numerator and denominator by cos θ. We get,

\(\frac{(4\, –\, tan\, θ)}{(2\, +\, tan\, θ)}\)

⇒ \(\frac{(4 – (4/3))}{(2 + (4/3))}\) [using the value of tan θ]

⇒ \(\frac{(12 – 4)}{(6 + 4)}\) [After taking LCM and cancelling it]

⇒ \(\frac{8}{10}\)

= \(\frac{4}{5}\)

\(\frac{4 cos\, θ\, -\, sin\, θ}{2 cos\, θ\, +\, sin\, θ}\) = \(\frac{4}{5}\)