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March 6, 2015

March 6, 2015

Posted by **McKenna Louise** on Friday, February 28, 2014 at 9:41pm.

1/(tanØ - secØ ) + 1/(tanØ + secØ) = -2tanØ

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

- Trig -
**Reiny**, Friday, February 28, 2014 at 10:06pmThat'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

- Trig -
**Steve**, Saturday, March 1, 2014 at 6:09amor, 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

- Trig -
**McKenna Louise**, Saturday, March 1, 2014 at 8:46amThanks to the both of you :)

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