# Trig

Prove the following:
1/(tanØ - secØ ) + 1/(tanØ + secØ) = -2tanØ

(1 - sinØ)/(1 + sinØ) = sec^2Ø - 2secØtanØ + tan^2Ø

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3. 👁 120
1. That's a better job of typing it.
I did the first of these in your previous post..

the 2nd:

LS = (1-sinØ)/(1+ sinØ)
= (1-sinØ)/(1+ sinØ) * (1-sinØ)/(1- sinØ)
= (1 - 2sinØ + sin^2 Ø)/(1 - sin^2 Ø)
= (1 - 2sinØ + sin^2 Ø)/cos^2 Ø
= 1/cos^2 Ø - 2sinØ/cos^2 Ø + sin^2 Ø/cos^2Ø
= sec^2 Ø - 2(sinØ/cosØ)*(1/cosØ) + tan^2 Ø
= sec^2 Ø - 2tanØ secØ + tan^2 Ø
= RS

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posted by Reiny
2. or, if you divide top and bottom by cosØ you have

(secØ-tanØ)/(secØ+tanØ)

now multiply top and bottom by (secØ-tanØ) and you have

(secØ-tanØ)^2 / (sec^2Ø-tan^2Ø)
= sec^2Ø - 2secØtanØ + tan^2Ø

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posted by Steve
3. Thanks to the both of you :)

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4. hj

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