[（1+cos20°）/（2sin20°）]-sin10°（cot5°-tan5°) Why do I see some people calculate = (root 3) / 2, and some people get tan10 / 2-2cos10? Well, well, it's better to have pictures or something. Otherwise, the semicolons are mixed up

[（1+cos20°）/（2sin20°）]-sin10°（cot5°-tan5°) Why do I see some people calculate = (root 3) / 2, and some people get tan10 / 2-2cos10? Well, well, it's better to have pictures or something. Otherwise, the semicolons are mixed up

Let's do it one by one,
(1+cos20°)/(2sin20°)=2cos²10°/(4sin10°cos10°)=cos10°/(2sin10°),
But cot 5 ° - Tan 5 °
=cos5°/sin5°-sin5°/cos5°
=(cos²5°-sin²5°)/(sin5°cos5°)
=cos10°/(1/2sin10°)
=2cos10°/sin10°,
So the original formula = cos10 ° / (2sin10 °) - 2cos10 °
= (cos10°-2sin20°)/(2sin10°)
=[cos10°-2sin(30°-10°)]/(2sin10°)
=(2cos30°sin10°))/(2sin10°)
=cos30°
=√3/2.

sin10°cos20°cos40°= ___ .

sin10°cos20°cos40°=sin20°cos20°cos40°
2cos10°=sin40° cos40° 
4cos10° =sin80° 
8cos10° =1
8，
So the answer is 1
Eight

Evaluation: evaluation: (1 + cos20 °) / (2sin20 °) - sin10 ° (Tan ^ - 15 ° - tan5 °) (solution process, don't use cot, I haven't learned it yet.)

Because (1 + cos 20 °) / (2 sin 20 °) = [1 + 2 (COS 10 °) ^ 2 - 1] / (4 sin 10 ° cos 10 °) = 2 (COS 10 °) ^ 2 / (4 sin 10 ° cos 10 °) = cos 10 ° (2 sin 10 °), sin 10 ° (1 / Tan 5 ° - Tan 5 °) = sin 10 ° (COS 5 ° / sin 5 °)

Evaluation: 1 + cos20 degree 2sin20°−sin10°(1 tan5°−tan5°)．

Original formula = 2cos210 degree
4sin10°cos10°−sin10° (cos5°
sin5°−sin5°
cos5°)=2cos210°
4sin10° cos10°− 2sin10° (cos25° −sin25°
2sin5°cos5°)
=cos10°
2sin10°−2cos10° ＝cos10° −2sin20°
2sin10°
=cos10°−2sin(30°−10°)
2sin10°＝cos10°−2sin30°cos10° +2cos30° sin10°
2sin10°
=cos30° ＝
Three
Two

Calculation: TAN1 °× tan2 °× tan3 °×... Tan88 °× tan89 °

tan89°=cot(90°-89°)=cot1°=1/tan1°
So TAN1 ° · tan89 ° = 1
In the same way
tan2°·tan88°=1
……
tan44°·tan46°=1
tan45°=1
So the original formula = 1

Tan 1 degree * Tan 2 degree * Tan 3 degree. Tan 87 Degree * Tan 88 Degree * Tan 89 degree

Tan 1 degree * Tan 2 degree * Tan 3 degree. Tan 87 Degree * Tan 88 Degree * Tan 89 degree
=Tan 1 degree * Tan 89 Degree * Tan 2 degree * Tan 88 Degree * Tan 3 degree * Tan 87 degree
=1×1×1×1.
=1
Note: Tan α × Tan (90 ° - α) = 1

Tan 1 degree * Tan 2 degree * Tan 3 degree *. * Tan 88 Degree * Tan 89 degree =?

tan1*tan89=1
tan2*tan88=1
tan45=1
So the result is. = 1

Confirmation: 3 + TAN1 ° · tan2 ° + tan2 ° · tan3 ° = tan3 ° tan1°．

It is proved that: 3 + TAN1 ° · tan2 ° + tan2 ° · tan3 °
=（1+tan1°•tan2°）+（1+tan2°•tan3°）+1
=tan2°−tan1°
tan(2−1)°+tan3°−tan2°
tan(3−2)°+1
=tan2°−tan1°+tan3°−tan2°
tan1°+1
=-1+tan3°
tan1°+1
=tan3°
tan1°
The original equation holds

The size relation of TAN1, tan2, tan3, tan4 And the size of sin1, sin2, SIN3, SiN4 The size relationship of Cos4, cos5pai ﹣ 4, sin7pai ﹣ 6

In Tan, only one or three quadrants are positive, and the others are negative
In sin, the first and second quadrants are positive and the others are negative
In COS, one and four quadrants are positive and the others are negative
It is suggested that you draw a coordinate diagram first, and then judge the position of an angle according to the coordinate diagram, which is close to the maximum value
tan2

tan1°tan2°+tan2°tan3°...+tan88°tan89°

∵tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)∴tanαtanβ=(tanα-tanβ)/tan(α-β) - 1∴tan1°tan2°+tan2°tan3°...+tan88°tan89°=(tan2°-tan1°)/tan1°+(tan3°-tan2°)/tan1°+...+(tan89°-tan88°)/tan1...