We need to prove that :

cosx/ (1-sinc) - sec x = tanx

We will start from the left side and prove the right side.

First, we know that sec x = 1/cosx

==> cosx / (1-sinx) - 1/cosx

==> Now we will rewrite using a common denominator cosx(1-sinx)

==> ( cosx*cosx - (1-sinx) / cosx(1-sinx)

==> (cos^2 x + sinx -1) / cosx(1-sinx)

==> We know that cos^2 x = 1- sin^2 x

==> (1-sin^2 x + sinx -1 ) / cosx(1-sinx)

==> Now we will factor:

==> (1-sinx)(1+sinx) - (1-sinx) / cosx(1-sinx)

We will factor (1-sinx)

==> (1-sinx)[ (1+sinx -1) / cosx(1-sinx)

Now we will reduce 1-sinx

==> sinx/ cosx = tanx...........q.e.d

Then we proved that cosx/ (1-sinx) - sec x = tanx.

We have to prove (cos x / (1 - sin x)) - sec x = tan x

use tan x = sin x / cos x and sec x = 1/ cos x

Start from the right hand side

(cos x / (1 - sin x)) - sec x

=> cos x/ (1 - sin x) - 1/cos x

=> (cos x)^2 - 1 - sin x / (cos x *(1 - sin x))

=> (-(sin x)^2 - sin x)/(cos x *(1 - sin x))

=> (-sin x)( sin x - 1)/(cos x *(1 - sin x))

=> (sin x)( 1 - sin x)/(cos x *(1 - sin x))

=> (sin x)/(cos x)

=> tan x

which is the right hand side.

This proves (cos x / (1 - sin x)) - sec x = tan x