It is known that Tan (α - β) = 2 / 1, Tan β = 3 / 1, and α ∈ (0, π), then α = the detailed process If the title is wrong, Tan (α - β) = 1 / 2, Tan β = 1 / 3, and α∈ (0, π), then α=

It is known that Tan (α - β) = 2 / 1, Tan β = 3 / 1, and α ∈ (0, π), then α = the detailed process If the title is wrong, Tan (α - β) = 1 / 2, Tan β = 1 / 3, and α∈ (0, π), then α=

tan(α-β)=(tanα-tanβ)/(1+tanα*tanβ)=1/2
(tanα-1/3)/(1+1/3tanα)=1/2
2(tanα-1/3)=1+1/3tanα
2tanα-2/3=1+1/3tanα
6tanα-2=3+tanα
5tanα=5
tanα=1
α=1/4π
(or α = 45 °)
Because Tan (α - β) = (Tan α - Tan β) / (1 + Tan α · Tan β)
That is, (Tan α - 1 / 3) / (1 + Tan α · 1 / 3) = 1 / 2
tanα-1/3=1/2+1/6tanα
tanα=5/6*6/5
So tan α = 1
α∈(0,π)
α=45°
Solve the equation 0.02 0.1 x-0.2-0.5 x + 1 = 3
(0.1x-0.2)/0.02-(x+1)/0.5=3
(5x-10)-(2x+2)=3
3x-12=3
3x=15
X=5
Divide both sides of the equation by 10,
(x-2)/2-(x+1)/5=0.3
Multiply both sides of the equation by 10
5x-10-2x-2=3
X=5
5x-20-50 / 100 (x + 1) = 300
5x-20-2（x+1）=300
5x-2x-2=300+20
3x=322
x=3/322
Three
Tan α = 3tan (α + β), β = π / 6, find the value of sin (2 α + β)
Expand Tan α = 3tan (α + β) and take β = π / 6 into it. Simplify Tan α = - root 3sin2 α = 2sin2 α cos α Cos2 α = cos & # 178; α - Sin & # 178; α because sin α and cos α are always positive and negative when Tan α is negative, so sin2 α = - root 3 / 2 Cos2 α = - 1 / 2, so sin (2 α + β) = sin2 α cos β
If we take B into the first formula, we can see that there are two sizes of a, and then we take the value of a - π / 3 or - π / 4 into 2A + B, and the solution is - 1 or negative half root sign 3.
(2 / 3 + 0.5) x = 4 / 5 solution equation
(2/3+1/2)X=4/5
7 / 6 times x = 4 / 5
X = 4 / 5 divided by 7 / 6 = 4 / 5 times 6 / 7 = 24 / 35
（2/3+1/2)X=4/5
2 / 3 and 1 / 2, get
(4/6+3/6)x=4/5
X = 4 / 5 divided by 7 / 6
x=24/35
If Tan α = 3tan (α + β) and β = π / 6, then sin (2 α - β) is obtained
tan(a+b)=tan(a+π/6)=(tana+√3/3)/(1-√3/3*tana)
So Tana = 3 (Tana + √ 3 / 3) / (1 - √ 3 / 3 * Tana)
3tana+√3=tana-√3/3*(tana)^2
√3/3*(tana)^2+2tana+√3=0
Multiply by √ 3 on both sides
(tana)^2+2√3tana+3=0
(tana+√3)^2=0
tana=-√3
aina/cosa=tana=-√3
sina=-√3cosa
(sina)^2+(cosa)^2=1
Get by substitution
(cosa)^2=1/4,(sina)^2=3/4
Because Sina / cosa = - √ 3
Solve the equation, 0.3 = 2x-1 x + 1
0.5 of 0.3 = 2x-1 of X + 1
03./0.5=(x+1)/(2x-1)
3/5=(x+1)/(2x-1)
5(x+1)=3(2x-1)
X=8
Find the value of 4cos ^ 2 (35 ') - cos (170') - Tan (160 ') sin (170')
The original formula is 2 (1 + cos70) + cos10 + tan20sin10
=2+2sin20+cos10+sin20sin10/cos20
=2+(2sin20cos20+(cos10cos20+sin20sin10))/cos20
=2+(sin40+cos10)/cos20
=2+(sin40+sin80)/cos20
=2+2sin60cos20/cos20
=2 + radical 3
>> 4*cos(35*pi/180)^2-cos(170*pi/180)-tan(160*pi/180)*sin(170*pi/180)
ans =
three point seven three two one
The solution equation of 0.5 parts x minus 3-0.2 parts x + 4 = 16
0.5 x minus 3-0.2 x + 4 = 16
J is 2 (x-3) - 5 (x + 4) = 16
2x-2*3-5x-5*4=16
5x-2x=-6-20-16
3x=-42
x=-14
Given that Tana = 1 / 3, tanb = - 2, 0 ° a < 90 ° and 270 ° B < 360 °, find the value of a + B
tan(a+b)=tana+tanb/(1-tanatanb)=-1
0°＜a＜90°,270°＜b＜360°,
be
270°
To solve the equation, 3 / 3 x = 0.5 plus 4 / 4 X
x/3=0.5+x/4
4x=6+3x
4x-3x=6
X=6