Prove:

1) 1 / sec X - tan X = sec X + tan X

2) cot A + tan A = sec A csc A

3)sec A - 1 / sec A + 1 = 1 - cos A / 1 + cos A

3 answers

I usually start these with changing everything to sines and cosines after I test them with some angle that I pick

#1. The question should say:
1/(secx - tanx) = secx + tanx
Those brackets are critical, the way you typed it, will not work

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

RS = 1/cosx + sinx/cosx
= (1+sinx)/cosx
= (1+sinx)/cosx * (1-sinx)/(1-sinx)
= (1 - sin^2 x)/(cosx(1-sinx)
= cos^2 x/(cosx(1-sinx))
= cosx/(1-sinx)
= LS
that was a tricky one!

#2 this one is easier, try it the way I did the first one

let me see your steps

#3, again, brackets are really important to establish the correct order of operation
(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)

I usually start with the more messy looking side
LS = (1/cosA - 1)/(1/cosA + 1)
= [ (1- cosA)/cosA ] / [ (1 + cosA)/cosA ]
= (1- cosA)/cosA * cosA/(1+cosA)
= (1 - cosA)/(1+cosA)
= RS

well, how about that ?
this particular sec/tan example falls out nicely:

1/(secx - tanx) = secx + tanx
1 = (secx+tanx)(secx-tanx)
1 = sec^2(x) - tan^2(x)

one of the fundamental identities, arising immediately from

sin^2 + cos^2 = 1

by dividing by cos^2
Purists will argue that to prove the validity of an identity, that is, prove that the equation is true, by using the rules of equations itself is not valid.