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.