(1-sin^6a-cos^6a)/(sin^2a-sin^4a)

(1-sin^6a-cos^6a)/(sin^2a-sin^4a)

Molecular = 1-sin ^ 6a-cos ^ 6A
=1-（sin^6a+cos^6a）
=1-（sin²a+cos²a)(sin^4 a-sin^acos6a+cos^4 a)
=1-(sin^4 a-sin^acos^a+cos^4 a)
=1-〔（sin²a+cos²a)²-3sin^acos^a〕
=1-（1-3sin²acos²a）
=3sin²acos²a
Denominator = (Sina + sin? A) (Sina sin? A)
=sina（1+sina）sina（1-sina）
=sin²a（1-sin²a）
=3sin²acos²a
Numerator / denominator = (3sin? ACOS? A) / (sin? ACOS? A) = 3

Simplification: Tan (π - α) · sin 2 (α + π / 2) · cos (2 π - α) / cos 3 (- α - π) · Tan (α - 2 π)

It can be reduced to:
tan（π-α)=-tana
sin²(α+π/2）=cos^2a
cos(2π-α）=cosa
cos³（-α-π)=-cos^3a
tan（α-2π）=tana
What you wrote is a bit confusing. You can't distinguish those in the numerator and those in the denominator. I didn't read it carefully just now,
Do you bring it in yourself or are you rewriting it

Simplify cos (A-3 π) / sin (a + 5 π) * sin (A-2 π) Simplify cos (A-3 π) / sin (a + 5 π) * sin (A-2 π) * cos (2 π - a)

cos(a-3π)/sin(a+5π)*sin(a-2π)*cos（2π-a）
=(-cosa)/(-sina)*sinacosa
=cos²a

Compare the big and small root 15 - root 13 with the 13 - root 11 inside the root sign. Pay attention to the second one

Nine

Root 1 and 9 / 7 equals? Root 4 / 16 × 25 equals? Root 3x / root 9x equals? 2 times root 5 / 10 equals? Help me again

Radical 1 and 7 / 9
=√ (16 of 9)
=4 / 3
Root number 4 / 16 × 25
=√(4×25)
=√4×√25
=2×5
=10
Radix 3x divided into Radix 9x
=√ (9x out of 3x)
=√3
Two times the root sign five times the root 10 equals
=2 √ (10 / 5)
=2 √ 2

Find the definite integral ∫ (upper limit is root 3, lower limit is 0) x times 1 + x square under root sign

Change yuan
Let x = tant DT = (sect) ^ 2DT
The integral limit becomes 0 to 60 degrees
The original formula = ∫ tantscet ^ 3DT = ∫ Sint / cost ^ 4dt
=-∫1/cost^4dcost
=1 / 3 * 1 / cost ^ 3 (0 to 60 degrees)
=7/3

Definite integral (1 to 0) radical xdx

Original = 1 / 2x ^ 2|1 to 0 = 1 / 2-0 = 1 / 2

2 SecA = sec (a + b) + sec (a-b), 0 ＜ a ＜ (π / 2) ＜ B ＜ π. It is proved that cosa = (radical 2) × cos (B / 2) 2secA=sec(A+B)+sec(A-B), 0＜A＜(π/2)＜B＜π, Confirmation: cosa = (radical 2) × cos (B / 2)

2 / cosa = 1 / cos (a + b) + 1 / cos (a-b)
2 / cosa = 2cosacosb / cos (a + b) cos (a-b) simplification
cosAcosAcosB=cosAcosAcosBcosB-sinAsinAsinBsinB
cosAcosA(1-cosB)=sinBsinB
cosAcosA=sinBsinB/(1-cosB)
Cosacosa = 1 + CoSb because [1 + CoSb = cos (B / 2) cos (B / 2)]
Cosa = (radical 2) × cos (B / 2)

Evaluation cos (arcsin Radix 5 / 4) to help do, to process. Thank you!

Let the angle a, known as Sina = 0.8, cos = positive and negative root (the square of 1-0.8) = plus or minus 0.6. Because the range of COS is from 0 to Pai (forgive me for not typing it), cosa = 0.6

What is one sixth of the root equal to

√6/6