Question

(secA + tanA-1)(secA-tanA+1)=2tanA

Answer 1

I assume you want to prove the identity

LS = (secA + tanA-1)(secA-tanA+1)=2tanA , expand it to get
= sec^2 A - secAtanA + secA + secAtanA - tan^2 A + tanA - secA + tanA - 1
= sec^2 A - tan^2 A + 2tanA - 1
= tan^2 A + 1 - tan^2 A + 2tanA - 1, since sec^2 A = tan^2 A + 1
= 2tanA
= RS