Proving a Trigonometric Identity

[LordofDirT]

Im supposed to verify that (1-sinx)/(1+sinx) = (secx-tanx)^2

RHS = (secx-tanx)^2 = (1/cosx - sinx/cosx)^2 = [(1-sinx) / cosx]^2

= [(1-sinx)(1-sinx)]/cosx^2 = (1-2sinx+sinx^2)/(1-sinx^2)

From here, I'm feeling pretty confused. I'm not even sure if all my values are correct.

 

[theUndergrad]

From there it's important to recognize that (1-a^2)=(1+a)(1-a) and then (1-2a-a^2)=(1-a)^2=(1-a)(1-a). Then it's just a matter of removing the common term.

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theUndergrad

http://www.theUndergraduateJournal.com

 

[LordofDirT]

ok,

starting from: [(1-sinx) / cosx]^2 = [(1-sinx)(1-sinx)]/cosx^2

using pythagorean identity

[(1-sinx)/(1-sinx)]/(1-sinx^2) = [(1-sinx)/(1-sinx)]/[(1-sinx)(1+sinx)] = (1-sinx)/(1+sinx) = RHS

Thanks.