If 4 tan θ = 3, evaluate `((4sin theta - cos theta + 1)/(4sin theta + cos theta - 1))`
Advertisement Remove all ads
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 = 3kand AB = 4k, 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 = − 5k , 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
Is there an error in this question or solution?
APPEARS IN
Advertisement Remove all ads