If Tana COTA = 2, then Tan ^ 2 + cot ^ 2A= If Tana COTA = 2, then how to calculate, ∵ Tana COTA = 2, ᙽ Tana COTA) ᙽ (Tana COTA) ^ 2 = 4,

If Tana COTA = 2, then Tan ^ 2 + cot ^ 2A= If Tana COTA = 2, then how to calculate, ∵ Tana COTA = 2, ᙽ Tana COTA) ᙽ (Tana COTA) ^ 2 = 4,

∵tana-cota=2,
∴(tana-cota)^2=4
tan²a -2tanacota+cot²a=4
tan²a-2+cot²a=4
∴tan²a+cot²a=6

The value of Ӟ cot Ӟ of Ӟ cot Ӟ a Ӟ cot Ӟ a Ӟ

(tanA)^2+(cotA)^2=6
(tanA-cotA)^2+2=6
(tanA-cotA)^2=4
|tanA-cotA|=2

How to calculate sin, cosin, Tan, cot, SEC and CSC? I can't recite from 0 ° to 360 ° one by one

You don't have to learn the first two if you don't know math
The rest can be done with the first two
tan=sin/cos
cot=cos/sin
sec=1/cos
csc=1/sec
You don't have to do it all. You just need to recite 0, 30, 60, 45
Other examinations are not taken

The result of simplifying [Tan (a + b) - Tana tanb] / [tanatan (a + b)] is

The original formula = [(Tana + tanb) / (1-tanatanb) - (Tana + tanb)] / [Tana (Tana + tanb) / (1-tanatanb)]
The numerator and denominator are divided by Tana + tanb
=[1/(1-tanatanb)-1]/[tana/(1-tanatanb)]
The numerator and denominator are multiplied by 1-tanatanb
=[1-(1-tanatanb)]/tana
=tanatanb/tana
=tanb

The function y = sin2x + 2sinxcosx-3cos2x, X ∈ R is known Sin2x cos2x are all squares

y=(1-cos2x)/2+sin2x-3(1+cos2x)/2
=sin2x-2cos2x-1
=√(1²+2²)sin(2x-a)+1
Where Tana = 2 / 1 = 2
=√6sin(2x-arctan2)+1

What is cot (- π - a) equal to or reduced to?

cot(-π-a)=-cot(π+a)=-cota

1-cos^6（α）-sin^6(α)/1-cos^4(α)-sin^4(α)

(1-cos^6α-sin^6α)/(1-cos^4α-sin^4α)
1=sin^2α+cos^2α
The original formula = (sin ^ 2 α + cos ^ 2 α - cos ^ 6 α - Sin ^ 6 α) / (sin ^ 2 α + cos ^ 2 α - cos ^ 4 α - Sin ^ 4 α)
Denominator = sin ^ 2 α (1-sin ^ 2 α) + cos ^ 2 α (1-cos ^ 2 α) = 2Sin ^ 2 α cos ^ 2 α
α - cosin ^ 4 (α - s) + α ^ 1
sin^2α(1-sin^4α)=sin^2α(1-sin^2α)(1+sin^2α)=sin^2αcos^2α(1+sin^2α)
Similarly, cos ^ 2 α (1-cos ^ 4 α) = sin ^ 2 α cos ^ 2 α (1 + cos ^ 2 α)
The original formula = sin ^ 2 α cos ^ 2 α (1 + sin ^ 2 α + 1 + cos ^ 2 α) / 2Sin ^ 2 α cos ^ 2 α = 3 / 2

A simplified problem of mathematics in Senior High School [1+2^(-1/32)][1+2^(-1/16)][1+2^(-1/8)][1+2^(-1/4)][1+2^(-1/2)]

The square difference formula [1-2 ^ (- 1 / 32)] [1 + 2 ^ (- 1 / 32)] [1 + 2 ^ (- 1 / 32)] [1 + 2 ^ (- 1 / 16)] [1 + 2 ^ (- 1 / 16)] [1 + 2 ^ (- 1 / 4)] [1 + 2 ^ (- 1 / 2)] = [1-2 ^ (- 1 / 16)] [1 + 2 ^ (- 1 / 16)] [1 + 2 ^ (- 1 / 4)] [1 + 2 ^ (- 1 / 4)] = [1-2 ^ (- 1 / 8)] [1 + 2 ^ (- 1 / 8)] [1 + 2 ^ (- 1 / 8)] [1 + 2 ^ (- 1 / 8)] [1 + 2 ^ (- 1 / 4)] [1 / 4] [1 + 2 ^ (- 1 / 4)] [1 / 4+ 2 ^ (- 1 / 2)

The value of (COS π / 5) (Cos2 π / 5) is equal to

Let's look at the fraction whose denominator is 1, (COS π / 5) (Cos2 π / 5) / 1, and then multiply sin (π / 5) at the same time. In this way, the formula of sin's double angle can be changed into: sin (2 π / 5) * cos (2 π / 5) / 2Sin (π / 5), and then the formula of sin's double angle can be used to obtain the formula of double angle of sin

Why is the value range of a function, that is, the number set B in its definition, wrong? If the function f (x) = | x + 1 | - | X-1 |, then f (x) is an odd function,

prove:
Because f (x) = |x + 1| - | x-1|,
f(-x)=|-x+1|-|-x-1|
=|-（x-1)|-|-(x+1)|
=|x-1|-|x+1|
=-f(x)
So f (x) is an odd function