Question

how does sec^4x - tan^4x

go with

sec^2x + tan^2x

Answer 1

sec^4x-tan^4x
=(sec^2x+tan^2x)(sec^2x-tan^2x)
= sec^2 x+ tan ^2 x as sec^2x -tan^2 x = 1

Answer 2

?????????????????

Answer 3

sec⁴x - tan⁴x = sec²x + tan²x →
1/(cos⁴x) - (sin⁴x)/(cos⁴x) = 1/(cos²x) + (sin²x)/(cos²x)
let us now multiply both sides by cos⁴x :
(cos⁴x)/(cos⁴x) - (sin⁴x)(cos⁴x) /(cos⁴x) =
(cos⁴x) /(cos²x) + (sin²x)(cos⁴x) /(cos²x) →
1 - sin⁴x = cos²x + sin²x cos²x


and let now express everything as a function of sinx:
1 - sin⁴x = 1 - sin²x + sin²x (1 - sin²x) →
1 - sin⁴x = 1 - sin²x + sin²x - sin⁴x →
1 - sin⁴x = 1 - sin⁴x (verified)

Answer 4

how did the negative in sec⁴x - tan⁴x become positive in sec²x + tan²x