Verification: (1 + SiNx + cosx) / (1 + SiNx cosx) - (1 + SiNx cosx) / (1 + SiNx + cosx) = 2 / TaNx

Verification: (1 + SiNx + cosx) / (1 + SiNx cosx) - (1 + SiNx cosx) / (1 + SiNx + cosx) = 2 / TaNx

(1+sinx-cosx)/(1+sinx+cosx) 1+sinx+cosx=1+2sinx/2*cosx/2+2(cosx/2)^2-1=2cosx/2*(sinx/2+cosx/2)1+sinx-cosx=1+2sinx/2*cosx/2-[1-2(sinx/2)^2=2sinx/2*(sinx/2+cosx/2)(1+sinx+cosx)/(1+sinx-cosx)=(cosx/2)/(s...

Verification: (TaNx * SiNx) / (TaNx SiNx) = (1 + cosx) / SiNx

TaNx = SiNx / cosx original formula = [(SiNx / Cox) * SiNx] / [(SiNx / cosx) - SiNx] = SiNx / (1-cosx) = [SiNx * (1 + cosx)] / [(1-cosx) * (1 + cosx)] = [SiNx * (1 + cosx)] / [SiNx * SiNx] = (1 + cosx) / SiNx

Given TaNx = - 3 / 4, find the square of cosx - the square of SiNx

tanx= -3/4
sinx/cosx= -3/4
sinx=(-3/4)cosx
(cosx)^2-(sinx)^2=(cosx)^2-[(-3/4)cosx]^2
=(cosx)^2-(9/16)(cosx)^2
=(7/16)(cosx)^2
=(7/16)/[1＋(tanx)^2]
=(7/16)/(1＋9/16)
=(7/16)/(25/16)
=7/25

SiNx cosx = sinxcosx. Then sin2x =?

2 (root 2-1)
Let sin X be m, cos X be n,
There is M-N = Mn; In addition: m ^ 2 + n ^ 2 = 1
Square of ① both sides: m ^ 2-2mn + n ^ 2 = (MN) ^ 2
1-2mn=(mn)^2
Sin2x = 2sinxcosx = 2Mn
That is to solve a quadratic equation of one variable

Given (SiNx + cosx) / (SiNx cosx) = 3, find TaNx, 2Sin ² x+(sinx-cosx) ² Value of

(sinx+cosx)/(sinx-cosx)=3
sinx+cosx=3sinx-3cosx
sinx=2cosx
tanx=sinx/cosx=2
sinx=2cosx
Bring in the identity sin ² x+cos ² x=1
cos ² x=1/5
sinxcosx=2cosxcosx=2cos ² x=2/5
sin ² x=1-1/5=4/5
Original formula = 2 × 4/5+sin ² x+cos ²- 2sinxcosx
=8/5+1-4/5
=9/5

Given TaNx = 2, calculate (1), 2cosx-3sinx / SiNx + cosx. (2), SiNx + cosx SiNx

tanx=sinx/cosx=2 sinx=2cosx1 (2cosx-3sinx)/(sinx+cosx)=(sinx-3sinx)/(sinx+sinx/2)=-2/(3/2)=-4/32 sinx+cosx-sinx=cosxsin ² x+cos ² x=1 3cos ² X = 1 cosx = ±√ 3 / 3 original formula = ±√ 3 / 3