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`.