cos^α-sin^α/1+2sinαcosα=1-tanα/1+tanα

cos^α-sin^α/1+2sinαcosα=1-tanα/1+tanα

(cos^α-sin^α)/(1+2sinαcosα)
=(cosα-sinα)(cosα+sinα)/(sinα+cosα)^2
=(COS α - sin α) / (sin α + cos α)
=(1-tanα)/(1+tanα)

It is proved that (1 + sin α) / (1 + sin α + cos α) = 1 / 2 * (1 + Tan α / 2)

Universal formula: Sina = (2tana / 2) / [1 + (Tana / 2) ^ 2]
cosa=[1－(tana/2)^2]/[1+(tana/2)^2]
Left = [1 + 2tana / 2 + (Tana / 2) ^ 2] / [2 + 2tana / 2]
=[(1+tana/2)^2]/2(1+tana/2)
=1/2*(1+tana/2)
=Right

It is proved that Tan β / 2 = sin β / (1 + sin β) = (1-cos β) / sin β

Tan β / 2 = sin β / (1 + sin β) = (1-cos β) / sin β. There is probably a problem with the question, (1 + sin β) should be 1 + cos β
tanβ/2=sinβ/（1+cosβ）=（1-cosβ）/sinβ
sinβ/（1+cosβ）=2sin（β/2）cos（β/2）/（1+2cos²（β/2）-1）
=2sin（β/2）cos（β/2）/（2cos²（β/2））
=sin（β/2）/cos（β/2）
=tan（β/2）
（1-cosβ）/sinβ =[1-(2cos²（β/2）-1）]/2sin（β/2）cos（β/2）
=2sin²（β/2）/[2sin（β/2）cos（β/2）]
==tan（β/2）

It is proved that Tan α / 2 = 1-cos or sin

prove:
1-cosɑ/sinɑ
={1-[1-2sin²(α/2)]}/[2sin(ɑ/2)cos(ɑ/2)]
=2sin²(ɑ/2)/[2sin(ɑ/2)cos(ɑ/2)]
=sin(ɑ/2)/cos(ɑ/2)
=tan(α/2)
∴ tan(α/2)=(1-cosɑ)/sinɑ

If Tan (π - α) = - 1 / 2, sin α cos α - 2Sin ^ 2A=

Because Tan (π - α) = - 1 / 2, that is π / 2

It is known that Tan α = 1 / 2. Ask for the help of sin α + cos α / 2Sin α - 3cos α sin α + 11cos α I want steps and details, not perfunctory

. you can draw a picture, ∵ Tan α = 1 / 2 ᙽ sin α = √ 5 / 5 cos α = 2 √ 5 / 5. Take it in and... Finally, it equals 49 / 5 - √ 5 / 25

The results showed that: (1-2sin α cos α) / (COS ^ α - Sin ^ α) = [(1-tan α) / (1 + Tan α)]

Conclusion: (1-2sin α cos α) / (COS ^ α - Sin ^ α) = [(sin α) ^ 2 + (COS α) ^ 2-2sin α cos α] / [(COS α - sin α) (COS α + sin α)] = [(sin α - cos α) ^ 2] / [(COS α - sin α) (COS α + sin α)] = (COS α - sin α) / (COS α + sin α) = (1-tan α) / (1 + tan)

Prove sin ^ 4 + cos ^ 4 = 1-1 / 2Sin ^ 2 (2 α)

sinα^4+cosα^4=(sinα^2+cosα^2)^2-2sinα^2cosα^2=1^2-2(sinα^2cosα^2)=1-1/2(sin2α)^2

It was proved that 1 + 2Sin α cos α / sin ^ 2 α - cos ^ 2 α = Tan α + 1 / Tan α - 1

The left = (sin? α + cos? α - 2Sin α cos α) / (sin α + cos α) (sin α - cos α) = (sin + cos α) / (sin α + cos α) (sin α - cos α) = (sin + cos α) / (sin α - cos α) divide cos α by cos α from sin α / cos α = Tan α, so left = (Tan α + 1) / (Tan α - 1)

It is proved that (sin α + cos α - 1) (sin α - cos α + 1) 2 sin α cos α = 1% sin α + cos α

(sin α + cos α - 1) (sin α - cos α + 1) 2 sin α cos α
=2sinacosa/[sina+(cosa-1)][sina-(cosa-1)]
=2sinacosa/[sin²a-cos²a+2cosa-1)
=2sinacosa/(2cosa-2cos²a)
=sina/(1-cosa)
=sina(1+cosa)/(1-cos²a）
=sina(1+cosa)/sin²a
=(1+cosa)/sina