# Prove the Trig identity: (cosx-sinx)/(cosx+sinx)= sec2x-tan2x Thanks in advance!

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Let us start with the RHS

`sec2x-tan2x`

`= 1/(cos2x)-(sin2x)/(cos2x)`

`= (1-sin2x)/(cos2x)`

We know the following trigonometric identities;

`sin^2x+cos^2x = 1`

`sin2x = 2sinxcosx`

`cos2x = cos^2x-sin^2x`

`RHS`

`sec2x-tan2x`

`= 1/cos(2x)-sin(2x)/(cos(2x))`

`= (1-sin2x)/(cos2x)`

`= (sin^2x+cos^2-sin2x)/cos2x`

`= (sin^2x-2sinxcosx+cos^2x)/(cos^2x-sin^2x)`

`= (sinx-cosx)^2/((cosx+sinx)(cosx-sinx))`

`= -(sinx-cosx)/(cosx+sinx)`

`= (cosx-sinx)/(cosx+sinx)`

`= LHS`

*So its is proved that;*

`(cosx-sinx)/(cosx+sinx) = sec2x-tan2x`

**Sources:**

`ERRRRATA- CCOORRRRIGE`

I did mistake to read topic, I'm sorry

`if`

`cosx -sinx/(cosx +sinx)= sec2x-tan2x`

`` isn't holds true, i.e. `x=pi/6`

`sqrt(3)/2- (1/2)/(1/2+sqrt(3)/2)=1/(1/2) -(sqrt(3)/2)/(1/2)`

`sqrt(3)/2-(1/2)/(1/2(1+sqrt(3)))=2-sqrt(3) `

`sqrt(3)/2-1/(1+sqrt(3))=2-sqrt(3)`

`(sqrt(3)(1+sqrt(3))-2)/(2(1+sqrt(3)))=2-sqrt(3)`

`(sqrt(3)+3-2)/(2(1+sqrt(3)))=2-sqrt(3)`

`(sqrt(3)+1)/(2(sqrt(3)+1))=2-sqrt(3)`

`1/2=2-sqrt(3)`

No way!

TO BE CONTINUE.......