Question

prove that sec4-sec2=tan4+tan2A

Answer 1

LHS=
${ \sec(x) }^{4} - { \sec(x) }^{2} = { ({ \sec(x) }^{2} })^{2} - { \sec(x) }^{2}$
ie
${(1 + { \tan(x) }^{2}) }^{2} - (1 + { \tan(x) }^{2} )$
ie
$1 + { \tan(x) }^{4} + 2 { \tan(x) }^{2} - 1 - { \tan(x) }^{2}$
ie
${ \tan(x) }^{4} - { \tan(x) }^{2}$
= RHS

Hence,
Proved