Simplifying Trig Identities using Factoring; Difficult Problem?

Question

I need to simplify this using factoring: sec3x – sec2x – secx +1

This is read as: Secant cubed x minus secant squared x minus secant x plus 1.

Need to get it to most reduced/simplified form USING FACTORING.

Thank you; very much appreciated.

I believe the answer should have tangent in the end.... I can get up to
sec^2x(secx-1) -(secx+1), but can someone please explain to me how to get (sec^2x-1)(secx-1) from that?

Answer 1

sec3x – sec2x – secx +1
= sec^2x(secx -1) -(secx-1)
= (sec^2x-1)(secx-1)
= (secx -1)(secx+1)(secx-1)

Answer 2

this is sort of an easy question if you know how to factor by grouping, if you dont know how to factor by grouping, you flat out cant do this problem

so just to clear up the problem a bit

let = (sec x) = x

so

the problem becomes

x^3 - x^2 - x + 1

now factor by grouping

x^2(x-1) - 1(x-1)

take x - 1 as a common factor and you get

(x^2 - 1)(x-1)

but (x^2 - 1) can be factored again by using DOTS, or difference of two squares,

(x^2 - 1)
factors into
(x+1)(x-1)

so the in the end we have
(x^2 - 1)(x-1) = (x+1)(x-1)(x-1)

remember we let sec x = x, so just go back into (x+1)(x-1)(x-1)
and replace all the x's with sec x

so the final equation is (secx +1)(secx -1)(secx -1)

by the x^2 = "x squared"

glad to help

Answer 3

im not sure if this is legal... but... this is what im speculate'n...

(sec^3(x) - sec^2(x)) + (-sec x + 1)
= sec^2(x)(sec x -1) - (sec x -1)
= (sec^2(x)+1) (sec x-1)