Solve #{ 2+2sin2x } /{ 2(1+sinx)(1-sinx) } =sec^2x+tanx# ?

Solve #{ 2+2sin2x } /{ 2(1+sinx)(1-sinx) } =sec^2x+tanx#

# 0 = { 2+2sin2x } /{ 2(1+sinx)(1-sinx) } - sec^2x - tanx#

# = { 2+2 (2 sin x cos x) } /{ 2(1-sin ^2 x) } - 1/cos^2x - sin x/cos x #

#= {1 + 2 sinx cos x}/{cos ^2 x} - 1/cos^2 x - {sin x cos x}/cos^2 x#

# = {sin x cos x}/{cos ^2 x} = tan x#

# tan x = 0 #

#x = k \pi quad# integer #k#

We have,

#(2+2sin2x)/(2(1+sinx)(1-sinx))=sec^2x+tanx#

#=>(2(1+sin2x))/(2(1-sin^2x))=sec^2x+tanx#

#=>(1+sin2x)/cos^2x=sec^2x+tanx#

#=>1+sin2x=sec^2xcos^2x+tanxcos^2x#

#=>1+sin2x=1+sinx/cosx xxcos^2x#

#=>sin2x=sinxcosx#

#=>2sin2x=2sinxcosx#

#=>2sin2x=sin2x#

#=>2sin2x-sin2x=0#

#=>color(red)(sin2x=0...to(A)#

#=>2x=kpi,kinZZ#

#=>x=(kpi)/2, kinZZ#
But, for this #x# ,#sinx=1=>1-sinx=0#
So,
#(2+2sin2x)/(2(1+sinx)(1-sinx))=(2+0)/(2(1+1)(0))=2/0to# undefined

Thus,
#x!=(kpi)/2, kinZZ#
Hence, there is no solution.!!
Again from #(A)#

#sin2x=0=>2sinxcosx=0=>sinxcosx=0#
#=>sinx=0 or cosx=0,where,tanx and secx# is defined.
#i.e. cosx!=0=>sinx=0=>color(violet)(x=kpi,kinZZ#
There is contradiction in result when we take #sin2x=0#.