（tanθ-1）/(sinθ-cosθ)=secθ

（tanθ-1）/(sinθ-cosθ)=secθ

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Let α be the third quadrant angle and cos α / 2 = - 1-cos ^ 2 [(π - α) / 2] under - radical, then α / 2 is the third quadrant angle

α is the third quadrant angle, which is 180 degrees

It is known that α is the second quadrant angle, β is the third quadrant angle, sin α = 3 / 5, cos β = - 3 / 5 (1) Calculate the values of cos α, sin β and sin (α + β) (2) Find the value of Tan (α + β)

(1) α is the second quadrant angle and β is the third quadrant angle
∴ cosα

If a is the third quadrant angle and sin (A / 2) < 0, find the quadrant where the angle (A / 2) is located

Because a is the third quadrant angle, the deduced a / 2 is the second and fourth quadrant angle
Because sin (A / 2) < 0, it means that (A / 2) is in the third or fourth quadrant
To sum up, a / 2 is in the fourth quadrant
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If α is the third quadrant angle and sin α / 2

α is the third quadrant angle
360k+180＜α＜360k+270
180k+90＜α/2＜180k+135
When k = 2n
360n + 90 ＜ α / 2 ＜ 360n + 135, α / 2 is the second quadrant angle
When k = 2n + 1
In the fourth quadrant, the ratio of the fourth quadrant to the third quadrant was 360 2
Sin α / 2 < 0
So alpha / 2 is in the fourth image

[sin (a-faction) cot (A-2)] / [cos (a-faction) Tan (A-2)]

Sin (a-faction) = - sin ACOS (a-faction) = - cos Acot a = cos A / sin atan a = sin a / cos a, so [sin (a-faction) cot (A-2)] / [cos (a-faction) Tan (A-2)] = - sin Acot A / (- cos atan a) = - cos A / (- sin a) = cot a

It is proved that: [sin α + cos (α + β) sin β] / [cos α - sin (α + β) sin β] = Tan (α + β)

Firstly, sin (α) = sin ((α + β) - β) = sin (α + β) cos (β) - cos (α + β) sin (β)
cos(α) = cos( (α+β) - β) = cos(α+β)cos(β) + sin(α+β) sin(β)
So the left = [sin ((α + β) - β) + cos (α + β) sin β] / [cos ((α + β) - β) - sin (α + β) sin β]
=Sin (α + β) cos (β) / cos (α + β) cos (β) = Tan (α + β) = right
END

It is proved that: (1 + Tan α + 1 / cos α) / (1-tan θ + 1 / cos α) = (1 + sin α) / cos α

(1+tanα+1/cosα)/(1-tanθ+1/cosα)=[(cosa+sina+1)/cosa]/[(cosa-sina+1)/cosa]=(1+sina+cosa)/(1-sina+cosa)∵(1-sina+cosa)(1+sina)-cosa(1+sina+cosa) =1-sin²a+cosa+sinacosa-cosa-sinacosa-cos²a ...

Given that a is an acute angle and COS (a + π / 6) = 4 / 5, the value of cosa is

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Given that sin (π + a) = 1 / 3 and a is the fourth quadrant angle, the values of COS (2 π + a), sin (a-5 π), Tan (A-7 π) are calculated

Sin (π + a) = 1 / 3
So Sina = - 1 / 3, cosa = √ (1-1 / 9) = 2 / 3 √ 2
And a is the fourth quadrant angle
So cos (2 π + a) = cosa = 2 / 3 √ 2
sin（a-5π）=-sin（5π-a）=-sin（π-a）=-sina=1/3
tan（a-7π）=-tan（7π-a）=tana=sina/cosa=（-1/3）/（2/3√2）=-√2/4