Answer :
tanx+secx-1/tanx-secx+1=secx+tanx is equivalent to 1+sinx/cosx.
Answer:
We have to prove that,
[tex]\frac{\tan x+ \sec x-1}{\tan x- \sec x + 1}=\secx+\tan x[/tex]
L.H.S.
[tex]\frac{\tan x+ \sec x-1}{\tan x- \sec x + 1}[/tex]
[tex]=\frac{\tan x+ \sec x-(\sec^2 x-\tan^2x)}{\tan x- \sec x + 1}[/tex]
( ∵ sec² x - tan² x = 1 )
[tex]=\frac{\tan x+ \sec x-(\sec x-\tan x)(\sec x+\tan x)}{\tan x- \sec x + 1}[/tex]
( ∵ a² - b² = ( a + b ) ( a - b ) )
[tex]=\frac{(\tan x+\sec x)(1-\sec x+\tan x)}{\tan x-\sec x + 1}[/tex]
[tex]=\tan x+\sec x[/tex]
= R.H.S
Hence, proved.....