1-sin2x/cosx=tan(π/4-x)

Answer 1

as written, it is not true. I think you meant
(1-sin2x)/cos2x
= (1 - 2sinx cosx)/(cos^2x - sin^2x)
= (cos^2x - 2sinx cosx + sin^2x)/(cos^2x - sin^2x)
= (cosx-sinx)^2 / ((cosx-sinx)(cosx+sinx))
= (cosx-sinx)/(cosx+sinx)
= (1-tanx)/(1+tanx)
= tan(π/4 - x)