Answer:
[tex]( \sec \theta - \tan \theta)( \sec \theta + \tan \theta)[/tex]
from difference of two squares:
[tex]{ \boxed{ \tt{( {a}^{2} - {b}^{2}) = (a - b)(a + b) }}}[/tex]
[tex] = { \sec }^{2} \theta - { \tan}^{2} \theta \\ = (1 + { \tan}^{2} \theta) - { \tan}^{2} \theta \\ = 1[/tex]
hence proved
