Question

What is `tan^4theta/(1-sin^2theta` in terms of non-exponential trigonometric functions?

Answer 1

`2/(cos 2x - 1)`

Explanation:

`f(x) = tan^4 x/(1 - sin^2 x) = (sin^4 x/cos^4 x)(1/cos^2 x)` =
`= (sin^4x/cos^6 x)`
Develop `sin^4 x` and `cos^6 x`

`sin^4 x = (sin^2 x)(sin^2 x) = (1/4)(1 - cos 2x)(1 - cos 2x) =`
`= (1/4)(cos 2x - 1)(cos 2x - 1)`

`cos^6 x = (cos^2 x)(cos^2 x)(cos^2 x) =`
`= (1/8)(cos 2x - 1)(cos 2x - 1)(cos 2x - 1)`
Finally,
`f(x) = [(1/4)(cos 2x - 1)^2]/[(1/8)(cos 2x - 1)^3] =`
`= 2/(cos 2x - 1).`