Verification 2 (1-sinx) (1 + cosx) = (1-sinx + cosx) ^ 2

Verification 2 (1-sinx) (1 + cosx) = (1-sinx + cosx) ^ 2

Left = 2 (1-sinx) (1 + cosx) = 2 (1 + cosx SiNx sinxcosx) = 1 + 1 + 2 (cosx SiNx) - 2sinxcosx = (SiNx) ^ 2 + (cosx) ^ 2-2sinxcosx + 2 (cosx SiNx) + 1 = (cosx SiNx) ^ 2 + 2 (cosx SiNx) + 1 = (cosx SiNx + 1) ^ 2 = (1-sinx + cosx) ^ 2 = right

Verification: 2sinx • cosx (sinx+cosx−1)(sinx−cosx+1)＝1+cosx sinx．

Left = 2sinx • cosx (SiNx + cosx − 1) (SiNx − cosx + 1) = 2sinx • cosxsin2x − (cosx − 1) 2 = 2sinx • cosxsin2x − cos2x + 2cosx − 1 = 2sinx • cosx − 2cos2x + 2cosx = sinx1 − cosx = SiNx (1 + cosx) (1 − cosx) (1 + cosx) = SiNx (1 + cosx) 1 − cos2x = 1 + co

Verify that cosx / (1 + SiNx) - SiNx / (1 + cosx) = 2 (cosx SiNx) / (1 + SiNx + cosx)

Left general score = (cosx + COS) ² x-sinx-sin ² x)/(1+sinx+cosx+sinxcosx)
=[cosx-sinx+(cosx+sinx)(cosx-sinx)]/(sin ² x+cos ² x+sinx+cosx+sinxcosx)
=2(cosx-sinx)(1+cosx+sinx)/(sin ² x+cos ² x+sin ² x+cos ² x+2sinx+2cosx+2sinxcosx)
=2(cosx-sinx)(1+cosx+sinx)/[1+(sinx+cosx) ²+ 2(sinx+cosx)]
=2(cosx-sinx)(1+cosx+sinx)/(1+sinx+cosx) ²
=2(cosx-sinx)/(1+cosx+sinx)
=Right
Proof of proposition

Verify that cosx / (1-sinx) = (1 + SiNx) / cosx Note that the numerator and denominator are enclosed in brackets

cosx/(1-sinx)=cosx(1+sinx)/(1+sinx)(1-sinx)=cosx(1+sinx)/1-sin^2x=cosx(1+sinx)/cos^2x=(1+sinx)/cosx
Skill: in the first formula, multiply the numerator and denominator by (1 + SiNx)

Verification: (1 + SiNx cosx) / (1 + SiNx + cosx) = Tan (x / 2)

prove:
(1+sinx-cosx)/(1+sinx+cosx)
= [(1-cosx)+sinx]/[(1+cosx)+sinx]
= {2*[sin(x/2)]^2+2sin(x/2)cos(x/2)}/{2*[cos(x/2)]^2+2sin(x/2)cos(x/2)}
= sin(x/2)/cos(x/2)
= tan(x/2)

Why (1-sinx) / cosx = Tan (x / 2)?

Wrong. (1 - SiNx) / cosx = [1 - Tan (x / 2)] / [1 + Tan (x / 2)]. It should be: (1 - cos x) / SiN x = Tan (x / 2), or: SiN x / (1 + cos x) = Tan (x / 2). = = = = = = = = 1. (1 - cos x) / SiN x = Tan (x / 2). Proof: because cos x = 1 -