How to prove 1-2sinxcosx / cos ^ 2 = 1-tanx / 1 + TaNx

How to prove 1-2sinxcosx / cos ^ 2 = 1-tanx / 1 + TaNx

The left numerator 1-2sinxcosx = SiNx ^ 2 + cosx ^ 2-2sinxcosx = (cosx SiNx) ^ 2 denominator cos2x = cosx ^ 2-sinx ^ 2 = (cosx SiNx) (SiNx + cosx) becomes (cosx SiNx) / (cosx + SiNx) 1-2sinxcosx / cos ^ 2x = (cosx SiNx) / (cosx + SiNx) {

Given TaNx = - 1 / 3, calculate 1 / 2 sinxcosx + cos ^ X

SiNx / cosx = TaNx = - 1 / 3cosx = - 3sinx, then cos? X = 9sin? X because sin? X + cos? X = 1, sin? X = 1 / 10cos? X = 9 / 10sinxcosx = SiNx (- 3sinx) = - 3sin? X = - 3 / 10, so the original formula = 3 / 4

It is proved that TaNx + Cotx = 2sinxcosx + sin cubic xsecx + cos cubic xcscx

tanx+cotx=sinx/cosx+cosx/sinx=(sin²x+cos²x)/sinxcosx=1/sinxcosx=(sin²x+cos²x)²/sinxcosx=(sin^4 x+2sin²xcos²x+cos^4 x)/sinxcosx=sin³xsecx+2sinxcosx+cos³xcsc...

Given TaNx = 3, find the value of 4sin ^ 2x + 3sinx * cosx + 6cos ^ X

snx/cosx=tanx=3
sinx=3cosx
Substitute sin? X + cos? X = 1
So cos? X = 1 / 10
sin²x=9/10
sinxcosx=(3cosx)cosx=3cos²x=3/10
So the original formula = 51 / 10

If TaNx = 2, then (1) 1 / 4sin x-3sinxcosx / cos x + 2sinxcosx

(1) At the same time, it can be divided by cos ^ 2x = (1 / 4) Tan ^ 2x ^ 2x-3tanx / cos ^ 2x + 2tanx = (1 / 4) Tan ^ 2x ^ 2x-3tanx / cos ^ 2x + 2tanx = (1 / 4) Tan ^ 2x-3tanx / [1 / (1 + Tan ^ 2x)] + 2tanx = (1 / 4) Tan ^ 2x-3sinx (1 + Tan ^ 2x) + 2tanx = (1 / 4) * 4-3 * 2 (1 + 4) + 4 = 1-30 + 4 + 4 = 1-30 + 4 + 4 = 1-30 + 4 = - 25 + 25 + 25 + 4 = - 25 + 25 + 25 + 25 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + it's a good idea

TaNx = 2, (1 + 2Sin ^ x) / (3sinxcosx-2cos ^ x) =?

(1 + sin? X) / (3sinxcosx cos? X) Yes, (1 + sin? X) / (3sinxcosx cos? X) = (sin? X + cos? X + sin? X) / (3sinxcosx cos? X) = (2tan? X + 1) / (3tanx-1)

Given TaNx = 2, find 5sinx ^ 2 + 3sinxcosx-2cosx ^ 2

Let TaNx = t = 2sin2x = 2T / (1 + T ^ 2) = 4 / 5cos2x = (1-T ^ 2) / (1 + T ^ 2) = - 3 / 55sinx ^ 2 + 3sinxcosx-2cosx ^ 2 = 3sinx ^ 2 + 1.5sin2x-2cos2x = 1.5 (1-cos2x) + 1.5sin2x-2cos2x = 1.5 + 1.5sin2x-3.5cos2x = 24 / 5

(3sinx + 5cosx) / (2cosx-3sinx) = 5, then TaNx =?

（3sinx+5cosx)/(2cosx-3sinx)=（3tanx+5）/（2-3tanx）=5
TaNx = 5 / 18

Given TaNx = 3, we calculate: (2cosx-3sinx) / (2sinx + 3cosx)

(2cosx-3sinx) / (2sinx + 3cosx)
=(2－3tanx)/(2tanx＋3)
=－7/9

TaNx = 2 sin square x + sinxcosx-2cos square x =?

(sinx)^2+sinxcosx-2(cosx)^2
=[(sinx)^2+(cos)^2]+sinxcosx-3(cosx)^2
=1+(cosx)^2 (sinx/cosx-3)
=1+(cosx)^2 (tanx-3)
=1-(cosx)^2
=(sinx)^2
(sincox) = (2) ^ 2
That is: (SiNx) ^ 2 = 4 (cosx) ^ 2
Because: (SiNx) ^ 2 + (COS) ^ 2 = 1
So: (SiNx) ^ 2 = 4 / 5