It is known that Tan (π / 4 - θ) + Tan (π / 4 + θ) = 4, and - π

It is known that Tan (π / 4 - θ) + Tan (π / 4 + θ) = 4, and - π

tan(pi/4+t)+tan(pi/4-t)=4 --->(1+tant)/(1-tant)+(1-tant)/(1+tant)=4 --->(1+tant)^2+(1-tant)^2=4(1+tant)(1-tant) --->2+2(tant)^2=4-4(tant)^2 --->6(tant)^2=2 --->(tant)^2=1/3 --->(cost)^2=1/[1+(tant)^2]...

Sin (2 α + β) + 2Sin β = 0: Tan α = 3tan (α + β)

sin(2α+β)+2sinβ=0
Sin (α + α + β) + 2Sin (a + β - α) = 0 (formula)
Sinacos (α + β) + cosasin (a + b) + 2Sin (α + β) cosa-2cos (α + β) Sina = 0 (expansion)
3cosasin(α+β)-sinacos(α+β)=0
3cosasin(α+β)=sinacos(α+β)
3cosasin(α+β)/cosacoa(α+β)=sinacos(α+β)/cosacoa(α+β)
tanα=3tan(α+β)

Given Tan α = 2, find sin α − 3cos α sin α + cos α=______ ．

Since Tan α = 2, sin α − 3cos α sin α + cos α = Tan α − 3tan α + 1 = 2 − 32 + 1 = - 13, so the answer is - 13

tan43°tan44°tan45°tan46°tan47°=?
tan43°tan44°tan45°tan46°tan47°
Fast D

Hello!
∵43°+47°=90°,44°+46°=90°
∴tan47°=cot43°,tan46°=cot44°
Original formula = tan43 ° cot43 ° tan44 ° cot44 ° tan45 ° = 1

Find the value of Sin & # 178; 42 ° - Tan 44 °× Tan 46 ° + cos & # 178; 42 ° and find the process (retain the root sign)

sin²42°— tan44°×tan46°+cos²42°
=sin²42°+cos²42°— tan44°×tan46°
=1— tan44°×tan46°
=1— tan44°×1/tan44°
=1-1=0

Sin ^ 2 32 ° + cos ^ 2 32 ° + Tan 44 ° * Tan 45 ° * Tan 46 ° = several processes should also be written

sin^2 32°+cos^2 32°+tan44°*tan45°*tan46°
Where sin ^ 2 32 ° + cos ^ 2 32 ° = 1
tan44°*tan45°*tan46°
= cot(90 -44) * tan45 * tan 46
= cot 46 * tan 46 * tan 45
= 1 * 1
= 1
So the original formula = 1 + 1 = 2

How should tan44 ° cot45 ° tan46 ° be calculated?
Is there a simple calculation? What
I've thought about it for a long time, but I can't remember what's easy?

tan44°=0.96568877480707404595802729970068
cot45°=1
tan46°=1.0355303137905695069588325512481

(1) Verification: sin 40 ° (Tan 10 ° √ 3) = - 1
(2) It is known that sin β + cos β = 1 / 5 and 0 < β < π. The values of sin β, cos β and Tan β are obtained

1.
tan10-√3
=sin10/cos10-√3
=(sin10-√3cos10)/cos10
=2sin(10-60)/cos10
=-2sin50/cos10
sin40(tan10-√3）
=-(2sin40sin50)/cos10
=-[cos(50-40)-cos(50+40)]/cos10
=-cos10/cos10
=-1
two
It is known that (sin β + cos β) & sup2; = 1 / 25
So sin β cos β = - 12 / 25
X 1 for sin β and x 2 for cos β
Then sin β and cos β are two solutions of the equation x ^ 2-1 / 5x-12 / 25 = 0
The equation is obtained according to Veda's theorem
x^2-1/5x-12/25=0
(x-4/5)(x+3/5)=0
x1=4/5 x2=-3/5
Because β∈ (0, π),
So sin β = 4 / 5, cos β = - 3 / 5
tanβ=sinβ/cosβ=-4/3

(High School Mathematics) rank Tan 1, Tan 2, Tan 3 and Tan 4 in descending order and explain the reasons

(1).tan4=tan[π+(4-π)]=tan(4-π).(2)tan3=tan[π-(π-3)]=-tan(π-3).(3)tan2=tan[π-(π-2)]=-tan(π-2).(4)0

tan10°tan30°+tan30°tan50°+tan50°tan10°

From Tan (a + b) = (Tana + tanb) / (1-tanatanb), Tana + tanb = Tan (a + b) (1-tanatanb) tan10 ° tan30 ° + tan30 ° tan50 ° + tan50 ° tan10 ° = tan30 ° (tan10 ° + tan50 °) + tan50 ° tan10 ° = tan30 ° tan60 ° (1-tan10 ° tan50 °) + tan50 ° tan10 °