Simplify (SiNx + TaNx) / cos ^ 2x + sin ^ 2x + cosx

Simplify (SiNx + TaNx) / cos ^ 2x + sin ^ 2x + cosx

(sinx+tanx)/(cos^2x+sin^2x+cosx)
=(sinx+sinx/cosx)/(1+cosx)
=sinx(cosx+1)/[cosx(1+cosx)]
=sinx/cosx
=tanx

Given sin (x-45 °) = quarter root 2, find sinxcosx, TaNx + 1 / TaNx

If sin (x-45 °) = √ 2 / 4 = sinxcos45 ° - cosxsin45 °, SiNx cosx = 0.5, and the square of both sides leads to 1-2sinxcosx = 0.25
sinxcosx=3／8.
tanx+1／tanx=(sin²x+cos²x)／(sinxcosx)=8／3.

It was proved that: (1-2sinxcosx) / (tanx-1) = 1 / (TaNx + Cotx) - cos ^ 2

Left = (sin? X-2sinxcosx + cos? X) / (SiNx / cosx-1)
=(sinx-cosx)^2/[(sinx-cosx)/cosx]
=cosx(sinx-cosx)
=sinxcosx-cos²x
Right = 1 / (SiNx / cosx + cosx / SiNx) - cos? X
=sinxcosx/(sin²x+cos²x)-cos²x
=sinxcosx/1-cos²x
=Sinxcosx cos? X = left
Proof of proposition

Given TaNx = 3, then 4sin'2x-3sinxcosx-5cos'2x =?

4sin'2x-3sinxcosx-5cos'2x
=8sinxcosx-3sinxcosx-10(1+cos^2 x)
=5tanx (1/1+tan^2 x)-10(1+1/(1+tan^2 x))
=5*3 /(1+9) -10 (1+1/10)
=1.5-11
=-9.5

Given TaNx = 2, find 2Sin ^ 2-3sinxcosx-2cos ^ 2x

Divide by sin ^ 2x + cos ^ 2x = 1
The numerator and denominator are divided by cos ^ 2x
So all the numbers become TaNx and 1

If 2cosx ^ 2 + 3sinxcosx-3sinx ^ 2 = 1, then TaNx= The answer is 1 or - 1 / 4,

The original formula = 1 + cos2x + (3 / 2) sin2x-3 / 2 + (3 / 2) cos2x = (3 / 2) sin2x + (5 / 2) cos2x-1 / 2 = 1
∴(3/2)sin2x+（5/2）cos2x=3/2
∴3sin2x+5cos2x=3
Let TaNx = a, from the universal formula sin2x = 2A / 1 + a? Cos2x = 2A / 1-A
Therefore, 6a / 1 + a? + 10A / 1-A? = 3
By solving the equation, a = TaNx = 1 or - 1 / 4

Given TaNx = 2, find the value of 4sinx-2cosx / 3cosx + 3sinx?

4sinx-2cosx/3cosx+3sinx
=[(4sinx-2cosx)/cosx]/[(3cosx+3sinx)/cosx]
=(4tanx-2)/(3tanx+3)
=(4*2-2)/(3*2+3)
=2/3

Given TaNx = 2, find the value of sin? X-3sinxcosx-1

Because sin? X + cos? X = 1
So the original formula = sin? X / (sin? X + cos? X) - 3sinxcosx / (sin? 2x + cos? X) - 1
Divide by cos? X
By SiNx / cosx = TaNx
Therefore, the original formula = tan? X / (tan? X + 1) - 3tanx / (tan? X + 1) - 1
=4/5-6/5-1
=-7/5

Given TaNx = 2, then sin ^ 2x + sinxcosx-2cos ^ 2x =?

The original formula = (sin? X + sinxcosx-2cos? X) / (sin? X + cos? X) the denominator is the same as the division of cos? X
=(tan? X + tanx-2) / (tan? X + 1) replace TaNx = 2 with
=4/5

If TaNx = 2, sin2x + 1 = () A. 0 B. 9 Five C. 4 Three D. 5 Three

∵tanx=2，∴sin2x+1=sin2x
sin2x+ cos2x+1=tan2x
tan2x+ 1+1=4
4+1+1=9
5，
Therefore, B