1-sinø/1=sinø=(secø-tanø)^2?

Question

prove i have no idea how to do these things i have more if your good at these...

ø is a greek letter i think but i means angle or "X"

Answer 1

is this correctly written?
ø=x
I suppose this is:
(1-sinx)/(1+sinx)=(secx-tanx)^2
LHS=[(1-sinx)(1-sinx)]/[(1+sinx)(1-sinx)]
=[(1-sinx)^2]/cos^2x
=[(1-sinx)/cosx]^2
=[1/cosx -sinx/cosx]^2
=(secx-tanx)^2
therefore LHS=RHS

Answer 2

sec(x)-tan(x)= 1/cos(x) - sin(x)/cos(x)
=(1-sin(x))/cos(x)
square this and you get (1-sin(x))^2/cos^2(x)
this equals (1-2sin(x)+sin^2(x))/cos^2(x)
I dont think that what you are trying to prove is actually true.