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Tan20 + tan40 + (radical 3) times tan20 times tan40
The first method: tan60 = Tan (20 + 40) = root number 3tan (20 + 40) = (tan20 + tan40) / (1-tan20tan40) = root 3tan20 + tan40 = root 3-root 3tan20tan40, so tan20 + tan40 + root 3 * tan20 * tan40 = root 3. The second method: tan60 = Tan (20 + 40) = (tan20 + tan40) / (1-tan
The value of Tan 20 ° + Tan 40 ° + root sign 3tan20 ° * Tan 40 °
tan20°+tan40°+√3tan20°*tan40°
=Tan (20 ° + 40 °) (1-tan20 ° tan40 °) (sum difference product formula) + √ 3tan20 ° tan40 °
=tan60°(1-tan20°tan40°)+√3tan20°tan40°
=√3(1-tan20°tan40°)+√3tan20°tan40°
=√3-√3tan20°tan40°+√3tan20°tan40°
=√3
tan20°+tan40°+ The value of 3tan20 ° and Tan 40 ° is () A. Three B. - Three C. Three Three D. - Three Three
tan20°+tan40°+
3tan20°•tan40°
=tan（20°+40°）[1-tan20°tan40°]+
3tan20°•tan40°
=
3[1-tan20°tan40°]+
3tan20°•tan40°
=
3-
3tan20°•tan40°+
3tan20°•tan40°
=
Three
So choose a
Calculate the values of (1) Tan 20 ° + Tan 40 ° + root 3 Tan 20 ° Tan 40 ° （2）（1+tan1°）（1+tan2°）… （1+tan44°）
tan60°＝(tan40°+tan20°)/(1-tan20°tan40°)
→√3-√3tan20°tan40°＝tan20°+tan40°
∴tan20°+tan40°+√3tan20°tan40°＝√3.
tan20°+tan40°+ The value of 3tan20 ° and Tan 40 ° is () A. Three B. - Three C. Three Three D. - Three Three
tan20°+tan40°+
3tan20°•tan40°
=tan（20°+40°）[1-tan20°tan40°]+
3tan20°•tan40°
=
3[1-tan20°tan40°]+
3tan20°•tan40°
=
3-
3tan20°•tan40°+
3tan20°•tan40°
=
Three
So choose a
If Tan α = (radical 3) / 3, then α=
Tan α = (root 3) / 3, then α = 30 ° I calculated it by calculator
Tan α - radical 3 ≥ 0 1 + Tan α ≥ 0 What is the solution set of α? Tan α - radical 3 ≥ 0; 1 + Tan α ≥ 0
Replace with a
tana>=√3=tan(π/3)=tan(kπ+π/3)
Tan increases in a period (K π - π / 2, K π + π / 2)
So K π + π / 3
Seeking Tan (how many + 60 degrees) = negative root 3
tan(x+60°)=-√3=tan(-60°)
x+60°=kπ-60°
X = k π - 120 ° K ∈ integer
If α + β = π / 3, and α and β are acute angles, then Tan α + Tan β + radical 3tan α Tan β =?
This problem uses Tan sum formula
tan(α+β)
=(tanα+tanβ)/(1-tanαtanβ)
=tanπ/3
=√3
∴tanα+tanβ=√3-√3tanαtanβ
tanα+tanβ+√3tanαtanβ=√3
Simplify Tan α + Tan (60 - α) + radical 3tan α Tan (60 - α)?
Replace with a
tan60=√3
tan(a+60-a)=√3
[tana+tan(60-a)]/[1-tanatan(60-a)]=√3
tana+tan(60-a)=√3-√3tanatan(60-a)
So Tana + Tan (60-a) + √ 3tanatan (60-a) = √ 3
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