If 4 tan θ = 3, evaluate `((4sin theta - cos theta + 1)/(4sin theta + cos theta - 1))`

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#### Solution

Given: 4 tan θ = 3 ⇒ tan θ = 3/4

Let us suppose a right angle triangle ABC right angled at B, with one of the acute angle θ. Let the sides be BC = 3*k*and AB = 4*k,* where k is a positive number

By Pythagoras theorem, we get

`AC^2 = BC^2 + AB^2`

`AC^2 = (3k)^2 + (4k)^2`

`AC^2 = 9k^2 + 16k^2`

`AC = sqrt(25k^2)`

`AC = +- 5k`

Ignoring AC = − 5*k* , as *k* is a positive number, we get

AC = 5k

if `tan theta = (BC)/(AB) = 3/4` then `sin theta = (BC)/(AC) = 3/5` and `cos theta = (AB)/(AC) = 4/5`

Putting the values in `((4 sin theta - cos theta + 1)/(4 sin theta + cos theta - 1))` we get

`((4xx3/5 - 4/5 + 1)/(4xx 3/5 + 4/5 -1)) = (((12- 4 + 5)/5)/((12 + 4 - 5)/5)) = 13/11`

Concept: Trigonometric Ratios

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