Calculation: 1-tan (5 π / 12) Tan (π / 4) ---- Tan (5 π / 12) + Tan (π / 4) That bar is the fractional line

Calculation: 1-tan (5 π / 12) Tan (π / 4) ---- Tan (5 π / 12) + Tan (π / 4) That bar is the fractional line

[1-tan（5π/12）tan(π/4）]/ [tan(5π/12）+tan(π/4）]
=1/tan(5π/12+π/4)
=1/tan(2π/3)
=-1/tan(π/3)
=-√3/3

Calculation of Tan π / 12-tan5 π / 12

tanπ/12=2-√3,tan5π/12=2+√3
Therefore, the original formula = - 2 √ 3

Calculate (Tan 10 ° - √ 3) cos 10 ° / sin 50 °

=2/sin 50°(1/2sin10°-√3/2cos10°)
=2/sin 50°(sin30sin10-cos30cos10)
=2/sin 50°(-cos40)
=-2

It is proved that the square of 1 + cos2a + 2 (Sina) is 2

Cos2a = square of cosa - square of sina
The left side of the original formula is equal to 1 + (COSA squared - Sina squared) + 2 (Sina) squared
=1 + cosa squared + Sina squared
=2

What condition does Sina = 1 / 2 be cos2a = 1 / 2

Sufficient and unnecessary conditions

The square 2-1 / - cos2a of reduction (Sina COSA)

[(sina-cosa)^2-1]/(-cos2a)
=-sin2a/(-cos2a)
=tan2a

If Tan α / 2 = 3, then 2sin2 α * cos α / (1 + Cos2 α) (1 + cos α)=

∵tanα/2=3
The formula of double angle is used briefly
sin2α=2sinαcosα,sinα=2sin(α/2)cos(α/2)
cos2α=2cos²α-1,1+cos2α=2cos²α
1+cosα=2cos²(α/2) 】
∴2sin2α*cosα/[(1+cos2α)(1+cosα)]
=2sinαcosα*cosα/[2cos²α(1+cosα)]
=sinα/(1+cosα)
=2sin(α/2)cos(α/2)/[2cos²(α/2)]
=sin(α/2)/cos(α/2)
=tan(α/2)=3

Tan α = 1 / 2, α∈ (π, 3 / 2 π), then cos (π / 2 + α)= Solution~~

=-sina
tana=sina/cosa=1/2
And Sina square + cosa square = 1
A is in the third quadrant
Sina = - root 5 / 5
Original formula = root 5 / 5

How is sin α + cos α / sin α - cos α equal to tan α + 1 / Tan α - 1,

(sin α + cos α) / (sin α - cos α)
=(tanα+1)/(1-tanα)

If sin θ cos θ = 1 2, then Tan θ + cos θ The value of sin θ is () A. -2 B. 2 C. ±2 D. 1 Two

Because sin θ cos θ = 1
2, then Tan θ + cos θ
sinθ=sinθ
cosθ+cosθ
sinθ=1
sinθcosθ=2．
Therefore, B is selected