1-cos2x sin2x / 1 cos2x sin2x = TaNx

1-cos2x sin2x / 1 cos2x sin2x = TaNx

It should be (1-cos2x + sin2x) / (1 + cos2x + sin2x) = TaNx
1-cos2x +sin2x=2sin^2x+2sinxcosx=2sinx(sinx+cosx),
1+cos2x +sin2x=2cos^2x+2sinxcosx=2cosx(sinx+cosx),
So (1-cos2x + sin2x) / (1 + cos2x + sin2x)
=2sinx(sinx+cosx)/2cosx(sinx+cosx)
=sinx/cosx=tanx.

How to simplify (1-tanx) (1 + sin2x + cos2x) In a functional form,

1-tanx=(cosx-sinx)/cosx
1+sin2x+cos2x
=(1+cos2x)+sin2x
=2cos^2x+2sinxcosx
=2cosx(sinx+cosx)
So the multiplication is equal to:
=[(cosx-sinx)/cosx]×[2cosx(sinx+cosx)]
=2[cos^2x-sin^2x]
=2cos2x

Verification: （1）1−2sinxcosx cos2x−sin2x=1−tanx 1+tanx； （2）（cosβ-1）2+sin2β=2-2cosβ．

(1) Left = 1 − 2sinxcosxcos2x − sin2x = cos2x + sin2x − 2sinxcosxs2x − sin2x = (cosx − SiNx) 2 (cosx + SiNx) (cosx − SiNx) = cosx − sinxcosx + SiNx = 1 − tanx1 + TaNx

If TaNx = 2, then (1 + sin2x) / cos2x= THX THX THX!

The original formula = (sin? X + cos? X + 2sinxcosx) / (COS? X-sin? X)
=(cosx+sinx)²/(cosx+sinx)(cosx-sinx)
=(cosx+sinx)/(cosx-sinx)
Divide cosx up and down
And SiNx / cosx = TaNx
=(1+tanx)/(1-tanx)
=-3

Let f (x) = AB, where vector a = (m, cos2x), B = (1 + sin2x, 1), X ∈ R, and the image of function y = f (x) passes through point (π / 4,2) (1) Find the value of real number M; (2) Find the minimum value of function f (x) and the set of X values at this time

(1).f(x)=ab=m(1+sin2x)+cos2x
=msin2x+cos2x+m
∵ function crossing point (π / 4,2)
∴msin(2×π/4)+cos(2×π/4)+m=2
m+m=2
M=1
(2).y=f(x)=sin2x+cos2x+1=√2sin(2x+45°)+1
When f (x) is the minimum
In other words, sin (2x + 45 °) = - 1
∴2x+π/4=2kπ-π/2
∴x=kπ-3π/8

Let f (x)= a• b. Where vector a=（m，cos2x）， B = (1 + sin2x, 1), X ∈ R, and the function y = f (x) passes through the point (π) 4，2) (I) find the value of real number M; (II) find the minimum value of function f (x) and the set of values of X

(I) ∵ f (x) = A.B = m (1 + sin2x) + cos2x = m + msin2x + cos2x is known as f (π 4) = m (1 + sin π 2) + cos π 2 = 2,  2m = 2, i.e. M = 1 (II) from (I), f (x) = 1 + sin2x + cos2x = 1 + 2Sin (2x + π 4)

Let f (x) = AB, where vector a = (m, cos2x), B = (1 + sin2x, 1), X belongs to the image of R and y = f (x), and the period of M and f (x) can be obtained by (π / 4,2) If M is a real number, the period should be the least positive period

f(x)=m+msin2x+cos2x
∵ over (π / 4,2)
∴2=m+msinπ/2+cosπ/2
2=m+m
M=1
f(x)=sin2x+cos2x+1
=√2sin(2x+π/4)+1
Minimum positive period: T = 2 π / 2 = π

Let f (x) = AB, where vector a = (m, cos2x), B = (1 + sin2x, 1), X belongs to the image of R and y = f (x), and the value range of FX is obtained through (π / 4,2)

f(x)=m(1+sin2x) + cos2x
SO 2 = m (1 + sin π / 2) + cos π / 2 = 2m, so m = 1
So f (x) = 1 + sin2x + cos2x = 1 + sin (2x + π / 4) * radical 2
Obviously, the range of sin (2x + π / 4) is [- 1,1], and that of F (x) is [1-radical 2,1 + radical 2]

Given the vector b = (m, sin2x), C = (cos2x, n), X ∈ R, f (x) = b * C, if the image of function f (x) passes through points (0,1) and (π/4,1). (1) Find the value of M, n; (2) Find the minimum positive period of F (x) and find the minimum value of F (x) on X ∈ [0, π / 4]; (3) When f (α / 2) = 1 / 5, α∈ [0, π], find the value of sin α Please write the detailed process, online, etc

(1)
f(x)
=b.c
= (m,sin2x).(cos2x,n)
= mcos2x+ nsin2x
f(0) = m = 1
f(π/4) = n = 1
(2)
f(x)= cos2x+sin2x
= √2(sin(2x+π/4))
Minimum positive period = π
min f(x) = f(0) = 1
(3)
f(α/2)=1/5
cosα+sinα = 1/5
(5cosα)^2 = (1- 5sinα)^2
25(sinα)^2 -5sinα-12 =0
(5sinα+3)(5sinα-4)=0
sinα =4/5 or sinα = -3/5 ( rejected )
ie sinα=4/5

In the plane rectangular coordinate system, a (2,0), B (0, 2), C (cos2x, sin2x), (0 < x < π) are known 2），f（x）= AB• AC (1) Find the minimum positive period of F (x) (2) Find the monotone increasing interval of F (x)

（1）f（x）=
AB•
AC=（-2，2）•（cos2x-2，sin2x）
=-2cos2x+4+2sin2x=4+2
2sin（2x-π
4），
Then the minimum positive period of F (x) is 2 π
2=π；
(2) Let 2K π - π
2≤2x-π
4≤2kπ+π
2，k∈Z，
Then K π - π
8≤x≤kπ+3π
8，
So the monotone increasing interval of F (x) is [K π - π
8，kπ+3π
8]，k∈Z．