Possible trigonometrical transformation?

tan = sin/cos, sec = 1/cos, sin^2 + cos^2 = 1

so substitute

(sin/cos + 1/cos - 1) / (sin/cos - 1/cos + 1) = LHS

Multiple LHS top and bottom by cos x, LHS =

(sin x + 1 - cos x) / (sin x - 1 + cos x) and show this equals (1 + sin x) / cos x

Cross multiple to show

(sin x + 1 - cos x) * cos x equals (1 + sin(x))(sin(x) - 1 + cos(x)), or

sin(x) cos(x) + cos(x) - cos^2(x) equals sin(x) + sin^2(x) -1 - sin(x) + cos(x) + sin(x)cos(x)

simplify to show (eliminating + sin(x)cos(x) from each side, and re-arrange RHS)

cos(x) - cos^2(x) equals [sin(x) - sin(x)] + sin^2(x) - 1 + cos(x) (so subtract cos(x) now from each side, giving)

-cos^2(x) = sin^2(x) - 1, rearrange

1 = sin^2(x) + cos^2(x) which is a truism - equation solved!!!