Evaluation: sin50 ° (1+ 3tan10°)=______ .

The original formula = sin 50 ° and COS 10 ° respectively+
3sin10°
cos10°=cos40°2sin40°
cos10°=sin80°
cos10°=sin10°
cos10°=1
So the answer is: 1

Why is 1 + root 3 * Tan 10 degrees equal to 1, It should be: (1 + radical 3 * tan10) * cos40

(1 + radical 3 * tan10) * cos40
=(1 + radical 3 * tan10) * cos40
=(1 + Radix 3sin10 / cos10) * cos40
=(cos10 + Radix 3sin10) * cos40 / cos10
=2sin(10+30)*cos40/cos10
=sin80/cos10
=sin80/sin80
=1

What is sin 40 ° multiplied by (Tan 10 ° - root 3)?

Sin40 ° multiplied by (tan10 ° - Radix 3) = sin40 ° * (sin10 ° / cos10 ° - sin60 °) = sin40 ° [(sin10 ° cos60 ° - cos10 ° sin60 °) / cos10 ° cos60 °] = sin40 ° * (- sin50 °) / (1 / 2 * cos10 °) = - 2sin40 ° sin50 ° / cos10 ° = - 2sin40 ° cos40

What are tan 20 degrees and Tan 10 degrees equal to

1. 2. Because Tana = Sina / cosa, the problem of finding tam10 degree is transformed into sin10 degree and cos10 degree problem! Since sin10 degree and cos10 degree only know one, they can use square relation to find the other. Therefore, in theory, it is OK to find one. For calculating sin10 degree and cos10 degree, we can

Simplified evaluation sin50 (1 + radical 3 * tan10) + tan10 + tan50 + Radix 3tan10tan50

=Sin50 * (root 3 * sin10 + cos10) / cos10 + root 3 * (1-tan10 * tan50) + root 3 * tan10 * tan50 = root 3 + sin50 * (1 + root 3tan10) = root 3 * (root 3 * sin10 + cos10) / cos10 = root 3 + sin100 / cos10 = root 3 + 1

First simplify and then evaluate: [1] the square of [1] A is - 4 / 2a-1 / 2, where a = root 3-2 【2】 The square of (3 root 5-2 root 7) 【3】 15 root sign 12 divided by 2 root sign 45 【4】 (3 and 1 / 2 of root - 5 and 1 / 3 of root) - (2 and 1 / 8 - root 20) 【6】 [5 root 2-6 root 3] [6 root 3 + 5 root 2] Thank you for your time

1. The original formula = 2A / (a + 2) (A-2) - 1 / (A-2) = (2a-a-2) / (a + 2) (A-2) (A-2) = 1 / (a + 2) when a = √ 3-2, the original formula = 1 / √ 3 = 3 / 3.2, the original formula = 43-12 √ 35.3, the original formula = 15 / 2 √ (4 / 9) = 15 / 2 × 2 / 3 = 5.4, the original formula = 3 √ 2 / 2-5 √ 3 / 3-3 / 3 / 2 / 2 + 2 √ 5 = 5.4, the original formula = 3 √ 2 / 2-5 √ 3 / 3 / 3 / 3 / 3 / 2 / 2 + 2 √ 5 = 5 √ 2-5 √ 3 √ / 3 + 2 √ 5.5, the original formula = (5 √ 2) ^

First simplify, then evaluate: A-2 / a square - 1 × (1 - (2a-3 of A-1)), where a = (root 2) - 1

The square of A-2 / A-1 × (1 - (2a-3 of A-1)) = - 1 / (a + 1)
When a = (radical 2) - 1
Original formula = - 2 / 2 root sign 2

First simplify and then evaluate a + B 1 + B 1 + a (a + b) Part B, where a = 2 / 2 root 5 + 1 b = 2 / 2 root 5-1

1/(A+B)+1/B+B/A(A+B)
=AB/AB(A+B)+A(A+B)/AB(A+B)+BB/AB(A+B)
=(AA+2AB+BB)/AB(A+B)
=(A+B)/AB
=1/A+1/B
=4 root sign 5

First simplify and then evaluate: (2a) a−1+a 1 − a) △ a, where a= 2+1．

The original formula = (2a)
a−1-a
a−1）÷a=a
a−1×1
A=1
a−1，
When a=
2 + 1,
Original formula = 1
2+1−1=1
2=
Two
2．

How many roots are equal to 18.75

Five times root three
Can you write

Evaluation: sin50 ° (1+ 3tan10°)=______ .