Simplification: (4 * Tan α) / [1 - (Tan α) ^ 2] specific steps are required, do not give the answer directly. Thank you

Simplification: (4 * Tan α) / [1 - (Tan α) ^ 2] specific steps are required, do not give the answer directly. Thank you

(4*tanα)/[1-(tanα)^2]
=2(2tanα)/[1-(tanα)^2]
=2tan(α+α)
=2tan(2α)

Simplify (1 + tan2 α) Cos2 α=______ .

（1+tan2α）cos2α=cos2α+sin2α
cos2α•cos2α=1，
So the answer is: 1

[1 / Tan A / 2-tan A / 2] * (1 + Tan a * Tan A / 2)

[1/tan(a/2)-tan(a/2)]×](1+tan a×tan(a/2)]={[cos²(a/2)-sin²(a/2)]/[cos(a/2)×sina/2]}×{[cosa×cos(a/2)+sina×sin(a/2)]/[cosa×cos(a/2)]}={2[1/2+cosa/2-1/2+cosa]/[2cos(a/2)×sina/2]}×{cos(...

Tan / Tan 1 / (Θ 2)

∵tanθ=2tan(θ/2)/[1-tan²(θ/2)]
∴1-tan²(θ/2)=2tan(θ/2)/tanθ
The original formula = [Tan 2 (θ / 2) - 1] / Tan (θ / 2)
=-2tan(θ/2)/tanθtan(θ/2)
=-2/tanθ
=-2cotθ

Simplification 1 + Tan α × Tan (α / 2)

1+tanα×tan（α/2)
= (cosαcos(α/2) + sinαsin(α/2))/cosαcos(α/2)
= (cos(α-α/2))/cosαcos(α/2)
= cos(α/2)/cosαcos(α/2)
= 1/cosα

2tan22.5°/1-tan^222.5°=

Solution;
2tan22.5°/1-tan^222.5°
=tan(2*22.5)
=tan45
=1

α + Tan (2) / (2)

It is known that Tan 2A = 2tana / [1 - (Tana) ^ 2] 1 / (Tan α / 2) - Tan (α / 2) = [1 - Tan (α / 2) ^ 2] / Tan (α / 2) = 2tan (A / 2) / [Tan a * Tan (α / 2) = 2 / Tan a = 2cos A / sin a1 + Tan ^ 2 α = 1 + sin ^ 2 α / cos ^ 2 α = 1 / cos ^ 2 α

The square of 1-tan22.5 degrees / tan22.5 degrees =?

The square of 1-tan22.5 degrees / tan22.5 degrees
=[(1-tan22.5 degree Square) * 2] / [tan22.5 Degree * 2]
=2

Given the vector a = (COS 75, sin 75), B = (COS 15, sin 15), then the angle between vector a and vector B is I haven't learned the sum and difference formula. Can I think of another way

Let a, B angle = x
|a|=|b|=1
a.b =|a||b|cosx
(cos75°,sin75°).(cos15°,sin15°) = cosx
cos75°cos15°+sin75°sin15°=cosx
cosx = cos60°
x = 60°

Given the vector a = (sin75, - cos75), B = (- cos15, sin15), then the modulus of a vector-b vector is

Given that the vector a = (sin75 °, - cos75 °), B = (- cos15 °, sin15 °), then: the module | vector a | = radical [sin | 75 ° + (- cos75 °)] = 1, | vector B | = radical [(- cos15 °)] = 1, and the scalar product vector a · B = sin75 ° (- cos15 °) + (- co