g(x) = sec^3(x) tan^4(x) in terms of only secant

Question

How do you write g(x) = sec³x tan⁴x in terms of only secant?

Answer 1

Break sec³(x)tan⁴(x) into sec³(x)tan²(x)tan²(x)

Use the Pythagorean Identity 1 + tan²(x) = sec²(x) but subtract 1 from both sides to yield:

tan²(x) = sec²(x) - 1

Substitute into the above to yield:

sec³(x)(sec²(x) - 1)(sec²(x) - 1)

so g(x) = sec³(x)(sec²(x) - 1)(sec²(x) - 1)

If you want you could multiply it out but I would advice against it.

Answer 2

sin^2(x) + cos^2(x) = 1

divide by cos^2(x) to get

tan^2(x) + 1 = sec^2(x)

tan^2(x) = sec^2(x)-1

square both sides

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

g(x) = sec^3(x)tan^4(x)

= sec^7(x) -2sec^5(x) + sec^3(x)