prove that sec⁴A - sec²A = tan⁴A +tan²A.....................

Sec⁴A - Sec²A

= Sec²A(Sec²A-1)

= (1+Tan²A)[(1+Tan²A) - 1] [Using: Sec²A-Tan²A=1 => Sec²A = 1+Tan²A]

= (1+Tan²A)Tan²A

= Tan²A + Tan⁴A

LHS = sec⁴A -sec²A

= (sec²A)² - sec²A

= (sec²A-secA) - (sec²A+secA) [ a²-b² = (a-b)(a+b) ]

= (1+tan²A - secA) (1+tan²A+secA)  [ sec²A =1 + tan²A ]

= 1 + tan²A + secA + tan²A + tan⁴A + tan²A.secA - secA - secA.tan²A - sec²A

=1 + tan²A + secA + tan²A + tan⁴A+ tan²A.secA - secA.tan²A - secA - sec²A

= 1 + tan²A + tan²A + tan⁴A-  sec²A

= 1 + tan²A + tan²A + tan⁴A - (1 + tan²A)

= 1 + tan²A + tan²A + tan⁴A - 1- tan²A

= tan²A + tan⁴A =  RHS

SOLVED !!!!!