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The simplified value tan70 ° cos10 ° + Radix 3sin10 ° tan70 ° - 2cos40 °
The original formula = 2tan70 ° [cos10 ° * 1 / 2 + (√ 3 / 2) sin10 °] - 2cos40 °
=2tan70°(cos10°sin30°+cos30°sin10°)-2cos40°
=2tan70°sin40°-2cos40°
=2cot20°sin40°-2cos40°
=2(cos20°/sin20°)2sin20°cos20°-2cos40°
=4(cos20°)^2-2[2(cos20°)^2-1]
=4(cos20°)^2-4(cos20°)^2+2
=2.
tan70°•cos10°（ 3tan20 ° - 1) is equal to () A. 1 B. 2 C. -1 D. -2
Tan70 ° cos10 ° (3tan20 ° - 1) = sin70 ° cos70 ° cos10 ° (3 · sin20 ° cos20 ° - 1) = cos20 ° cos10 ° sin20 ° - 3sin20 ° cos20 ° = cos10 ° sin20 °× 2Sin (20 ° - 30 °) = - sin20 ° sin20 ° = - 1
Evaluation: 1−2sin10°cos10° cos10°− 1−cos2170°．
Original formula=
1−2sin10°cos10°
cos10°−
1−cos2170°
=
sin210°−2sin10°cos10°+cos210°
cos10°−
sin2170°
=|sin10°−cos10°|
cos10°−|sin170°|
=cos10°−sin10°
cos10°−|sin(180°−10°)|
=cos10°−sin10°
cos10°−sin10°=1
Simplification 1−2sin10°cos10° sin170°− 1−sin2170°=______ .
1−2sin10°cos10°
sin170°−
1−sin2170°
=
(sin10°−cos10°)2
sin10°−
cos210°
=cos10°−sin10°
sin10°−cos10°
=-1．
So the answer is: - 1
tan70°•cos10°（ 3tan20 ° - 1) is equal to () A. 1 B. 2 C. -1 D. -2
Tan70 ° cos10 ° (3tan20 ° - 1) = sin70 ° cos70 ° cos10 ° (3 · sin20 ° cos20 ° - 1) = cos20 ° cos10 ° sin20 ° - 3sin20 ° cos20 ° = cos10 ° sin20 °× 2Sin (20 ° - 30 °) = - sin20 ° sin20 ° = - 1
tan70°•cos10°（ 3tan20 ° - 1) is equal to () A. 1 B. 2 C. -1 D. -2
Tan70 ° cos10 ° (3tan20 ° - 1) = sin70 ° cos70 ° cos10 ° (3 · sin20 ° cos20 ° - 1) = cos20 ° cos10 ° sin20 ° - 3sin20 ° cos20 ° = cos10 ° sin20 °× 2Sin (20 ° - 30 °) = - sin20 ° sin20 ° = - 1
tan70°•cos10°（ 3tan20 ° - 1) is equal to () A. 1 B. 2 C. -1 D. -2
Tan70 ° cos10 ° (3tan20 ° - 1) = sin70 ° cos70 ° cos10 ° (3 · sin20 ° cos20 ° - 1) = cos20 ° cos10 ° sin20 ° - 3sin20 ° cos20 ° = cos10 ° sin20 °× 2Sin (20 ° - 30 °) = - sin20 ° sin20 ° = - 1
tan70°•cos10°（ 3tan20 ° - 1) is equal to () A. 1 B. 2 C. -1 D. -2
tan70°•cos10°（
3tan20°-1）
=sin70°
cos70°•cos10°（
3•sin20°
cos20°-1）
=cos20°cos10°
sin20°•
3sin20°−cos20°
cos20°
=cos10°
sin20°×2sin（20°-30°）
=−sin20°
sin20°
=-1．
Therefore, C
How to simplify the radical 1 + 2sin10 ° cos10 '
Let's start with the root sign,
1+2sin10°cos10°
=（sin10°)^2+(cos10°)^2+2sin10°cos10°
=(sin10°+cos10°)^2
Then the square root and the square are cancelled at the same time to obtain sin 10 ° + cos 10 °
Evaluation: 1−2sin10°cos10° cos10°− 1−cos2170°．
The original formula is as follows: 1 − 2sin10 ° cos10 ° − 1 − cos2170 ° = sin210 ° - 2sin10 ° cos10 ° + cos210 ° cos10 ° - sin2170 ° = | sin 10 ° - cos10 ° | sin 170 ° = cos10 ° - Sin 10 ° cos10 ° − sin (180 ° - 10 °) | = cos10 ° - Sin 10 ° cos10 ° − sin (180 ° - 10 °)
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