Let α be the largest angle of an obtuse triangle. If Tan (α + π / 4) = 1 / 2, then sin α + cos α=

Let α be the largest angle of an obtuse triangle. If Tan (α + π / 4) = 1 / 2, then sin α + cos α=

Let Tan α = x, then (1 + x) / (1-x) = 1 / 2, then x = - 1 / 3
If α is obtuse angle, sin α > 0, cos α 0, then cos α = - 3K

Given sin α + cos β = √ 3 / 2 cos α + sin β = √ 2, find the value of Tan α cot β

Replace α, β with a and B
Sina + CoSb = radical 3 / 2
The square is:
sin^2a+2sinacosb+cos^2b=3/4.(1)
Cosa + SINB = radical 2
The square is:
cos^2a+2sinbcosa+sin^2b=2.(2)
(1)+(2):
1+2(sinacosb+sinbcosa)+1=11/4
2 sin (a + b) = 3 / 4
sin(a+b)=3/8
tana*cotb
=sina/cosa*cosb/sinb
=(sinacosb)/(cosasinb)

If a ∈ (π / 2,3 / 2 π), Tan (α - 7 π) = - 3 / 4, then the value of sin α + cos α is equal to

Because, Tan (α - 7 π) = - 3 / 4, namely: Tan [6 π + (π - α)] = 3 / 4
So, Tan α = - 3 / 4

If Tan θ = 2, then sin (π 2+θ)−cos(π−θ) sin(π 2−θ)−sin(π−θ)=______ .

∵sin（π
2±θ）=cosθ，cos（π-θ）=-cosθ，sin（π-θ）=sinθ
The original formula = cos θ + cos θ
cosθ−sinθ=2
1−sinθ
cosθ=2
1−tanθ=2
1−2=-2
So the answer is: - 2

Given Tan α = 3, what is sin α cos α equal to

With the help of the solution sin ^ 2 + cos ^ 2 = 1, sinacosa = sinacosa / (sin ^ 2 + cos ^ 2) = (sinacosa / cosa ^ 2) / (sin ^ 2 / cos ^ 2 + cosa ^ 2 / cosa ^ 2)
=tana/(1+tan^2)=3/10

Sin α is known 2+cosα 2= Three If cos α is less than 0, then Tan α is equal to () A. Two Two B. - Two Two C. 2 Five Five D. -2 Five Five

Square the left and right sides of the known equation to obtain: (sin α)
2+cosα
2）2=1
3，
That is, 1 + sin α = 1
3, sin α = - 2
3，
∵cosα＜0，∴cosα=-
1-sin2α=-
Five
3，
Then Tan α = sin α
cosα=2
Five
3．
Therefore, C is selected

If Tan α = - 2 / 3, sin α cos α is equal to

sinαcosα=sin2α
Tan α + cot α = 2 / sin2 α
-2/3-3/2=-（4/6+9/6）=-13/6=2/sin2α
sin2α=-2*6/13=-12/13

sin(-a)sin【5/(2-a)】tan(a-2π)/cos【a-（π/2）】cos【a-（3π/2）】-tan（π-a）tan【（3π/2）-a】 Tan α = 1 / 3 reduced evaluation {sin (- a) sin [(5 / 2) - A] Tan (A-2 π)} / {cos [a - (π / 2)] cos [a - (3 π / 2)] - Tan (π - a) Tan [(3 π / 2) - A]} sorry, it should be

sin(-a)sin【5/(2-a)】tan(a-2π)/cos【a-（π/2）】cos【a-（3π/2）】-tan（π-a）tan【（3π/2）-a】=-sin(a)sin【5/(2-a)】tan(a)/cos【（π/2）-a】cos【（3π/2）-a】-tan(a)tan(π/2-a)
=-sin(a)sin【5/(2-a)】tan(a)/sin(a)cos(a)-tan(a)cot(a)
=-sin【5/(2-a)】tan(a)/cos(a)-tan(a)cot(a)

It is known that sin β = 3.5, (90

cosb=-4/5, cosa=sin(a+b)=sina*cosb+cosa*sinb=-4/5 sina+3/5 cosa ,sina/cosa=-1/2,∴tan(a+b)=sin(a+b)/cos(a+b)=cosa/(cosa*cosb-sina*sinb) =cosa/(-4/5 cosa-3/5 sina) = -5/(4+3sin...

Given sin (π + a) = 3 / 5, find cos (π - a) Tan (- a)

cos(π-a)tan(-a)=(-cosa)(-tana)=cosa*tana=sina
sin(π+a)=3/5
sina=-3/5
cos(π-a)tan(-a)=sina=-3/5