verify the identity (tan x – sec x + 1)/(tan x + sec x - 1) = cos x(1 + sin x)?

as sec^2 x = tan ^2 x + 1 so 1 = sec^2 x - tan ^2 x

so putting 1 = sec^2 x- tan ^2 x in denominator we get

LHS -= (tan x – sec x + 1)/(tan x + sec x - 1)
= (tan x – sec x + sec^2 x- tan ^2 x)/(tan x + sec x - 1) replacing 1 with sec^2 x - tan ^2 x in numerator
= (tan x – sec x + (sec x- tan x)( sec x + tan x))/(tan x + sec x - 1) factoring sec^2 x - tan ^2 x
= (sec x- tan x)(-1 + sec x + tan x)/(tan x + sec x -1) taking sec x - tan x common
= sec x - tan x canceling common part
= 1/cos x - sin x/ cos x
= (1- sin x)/cos x
= cos x(1- sin x)/cos^2 x( multiplying numerator and denominator by cos x)
= cos x(1- sin x)/(1- sin ^2 x)
= cos x(1- sin x)/(1- sin x )(1+ sin x)
= cos x/(1+ sin x)
proved

Prove cos x /(1+sin x) + ((1+sin x )/cos x ) = 2 sec x

cos x/(1+sin x) + (1+sin x/cos x
= cos x(1- sin x)/((1+ sin x)(1- sin x)) + (1+sin x)/cosx
= cos x (1- sin x)/(1- sin ^x) + (1+sin x)/cos x
= cos x(1- sin x)/cos^2 x) + (1+sin x)/cos x
= (1- sin x)/cos x + (1+sin x)/cos x
= (1- sin x + 1 + sin x)/ cos x
= 2/ cos x
= 2 sec x