Solve the equation `secx=1+tanx`?

`secx=1+tanx` can be written as

`1/cosx=1+sinx/cosx`

and assuming `cosx!=0` this

`1=cosx+sinx`

or `sqrt2(sinxcos(pi/4)+cosxsin(pi/4))=1`

or `sqrt2sin(x+pi/4)=1`

or `sin(x+pi/4)=1/sqrt2=sin(pi/4)`

Hence `x+pi/4=npi+(-1)^n(pi/4)`

or `x=npi+(-1)^n(pi/4)-pi/4`

which simplifies to `x=2npi` or `x=2npi+pi/2`

Equation given

`secx=1+tanx`

`=>secx-tanx=1.....(1)`

Again we know

`sec^2x-tan^2x=1....(2)`

Dividing (2) by (1) we get

`secx+tanx=1....(3)`

Adding (1) and (3) we get

`2secx =2`

`=>cosx=1`

`=>x=2npi" where " n in ZZ`

Putting `cosx=1` in equation (1) we get
`tanx=0`

`=>x=npi" where " n in ZZ`

but for n odd `cosx=cosnpi=-1`

So the only solution is `x=2npi`

`1/(cos x) = 1 + sin x/(cos x)`
`1 = sin x + cos x` (condition `cos x != 0`)
`sin x + cos x = sqrt2cos (x - pi/4) = 1`
`cos (x - pi/4) = 1/sqrt2 = sqrt2/2`
Trig table and unit circle give 2 solutions for `(x - pi/4)`-->
`x - pi/4 = +- pi/4`
a. `x - pi/4 = pi/4 + 2kpi`
`x = pi/4 + pi/4 = pi/2 + 2kpi`
This answer is rejected because of the above condition:
`(cos x !=`0 --> `x != pi/2, and x != (3pi)/2)`
b. `x - pi/4 = - pi/4 + 2kpi`
`x = 2kpi`
Check.
`x = 2pi`--< cos x = 1 --> sec x = 1/1 = 1 --> tan x = 0
1 + tan x = 1 + 0 = 1. Proved.