Question

Prove(sec theta-tan theta)(sec theta+tan theta)=1​

Answer 1

Answer:

$( \sec \theta - \tan \theta)( \sec \theta + \tan \theta)$

from difference of two squares:

${ \boxed{ \tt{( {a}^{2} - {b}^{2}) = (a - b)(a + b) }}}$

$= { \sec }^{2} \theta - { \tan}^{2} \theta \\ = (1 + { \tan}^{2} \theta) - { \tan}^{2} \theta \\ = 1$

hence proved