It is known that Tan α - Tan β = 2tan α ^ 2tan β, and α β is not equal to K π / 2. Try to find sin β sin (2 α + β) / sin β Or it can give you ideas This is known: Tan α - Tan β = 2tan α Tan α Tan β

It is known that Tan α - Tan β = 2tan α ^ 2tan β, and α β is not equal to K π / 2. Try to find sin β sin (2 α + β) / sin β Or it can give you ideas This is known: Tan α - Tan β = 2tan α Tan α Tan β

tanα-tanβ=2tan^2αtanβ
tanα=tanβ(2tan^2α+1)
sinαcosα=tanβ(2sinα^2+cos^2α)
sin2α=2tanβ(1+sin^2α)=2tanβ+2sin^2αtanβ
sin2α+cos2αtanβ=3tanβ
sin2αcosβ+cos2αsinβ=3sinβ
sin(2α+β)=3sinβ
So sin (2 α + β) / sin β = 3

Known Tan α - Tan β = 2tan ^ 2 α Tan β, and α, β are not equal to K π / 2 (K ∈ z), try to find the value of sin (2 α + β) / sin β. Please!

∵tanα-tanβ=2（tanα）^2 tanβ
The common factor is extracted and divided by the same factor, and the result is: Tan β = Tan α / (2tan ^ 2 α + 1)
∵tan^2α=sec^2α-1=-1+(1/cos^2α)
∴sin(2α+β)/sinβ=(sin2α*cosβ+cos2α*sinβ)/sinβ
=（sin2α*（2tan^2α+1）/tanα）+cos2α
=4-2cos^2α+2cos^2α-1
=3

1+cos2θ+2sin²θ=2 1-cos2θ/1+cos2θ=tan²θ

1+cos2θ+2sin²θ=1+cos2θ+1-(1-2sin²θ)=2+cos2θ-cos2θ=2
(1-cos2θ)/(1+cos2θ)=(1-(1-2sin²θ))/(1+(2cos²θ-1))=2sin²θ/(2cos²θ)=tan²θ

If Tan θ = 3, then 2Sin θ + cos θ / 2Sin θ - cos θ=
We have to go through it in detail

According to tan θ = 2Sin θ / cos θ, the numerator denominator divided by cos θ at the same time is changed into: 2Sin θ + cos θ / 2Sin θ - cos θ = 2tan θ + 1 / 2tan θ - 1 = 7 / 5