Question

If 3tan θ   4 , show that `((4cos theta - sin theta ))/((4 cos theta + sin theta))=4/5`

Answer 1

Let us consider a right ΔABC right angled at B and ∠𝐶 = 𝜃.
We know that tan 𝜃 =`(AB)/(BC) = 4/3`

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
`AC^2 = AB^2 + BC^2`
`⟹ AC^2 = 16K^2 + 9K^2`
`⟹ AC^2 = 25K^2`
⟹ AC = 5k
Now, we have:

`sin theta = (AB)/(AC) = (4K)/(5K)=4/5`

`Cos theta = (BC)/(AC) = (3K)/(5K)=3/5`

Substituting these values in the given expression, we get:

`(4 cos theta - sin theta)/(2 cos theta + sin theta)`

`= (4(3/5)-4/5)/(2 (3/5)+4/5)`

`= (12/5-4/5)/(6/5+4/5)`

`= ((12-4)/5)/((6+4)/5)`

`= 8/10 = 4/5 = RHS`

i.e., LHS = RHS
Hence proved.