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Math Help - Simpiifying trig functions...urgent help needed

  1. #1
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    Simpiifying trig functions...urgent help needed

    I have tried these several times and cannot figure out how to simplify these...
    There are 2 problems...I would really appreciate it if someone would look them over and guide me...I am having a very hard time with this unit
    Thank you
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  2. #2
    MHF Contributor red_dog's Avatar
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    (1+\sin x)(1-\sin x)=1-\sin^2x=\cos^2x


    \displaystyle\cot^2y(\sec^2y-1)=\frac{\cos^2y}{\sin^2y}\left(\frac{1}{\cos^2y}-1\right)=\frac{\cos^2y}{\sin^2y}\cdot\frac{1-\cos^2y}{\cos^2y}=\frac{\sin^2y}{\sin^2y}=1
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  3. #3
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    Hello, aikenfan!

    If you're familiar with basic identities, these are simple . . .


    (10) Prove: . \cot^2\!y(\sec^2\!y - 1) \;=\;1

    \text{We have: }\;\cot^2\!y\underbrace{(\sec^2\!y - 1)}_{\text{This is }\tan^2\!y} \;=\;\cot^2\!y\cdot\tan^2\!y \;=\;(\cot y\cdot\tan y)^2 \;=\;(1)^2 \;=\;1

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  4. #4
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    Thank you very much for your help...I have been having quite a hard time with these...I know the identities, but I am never sure which way to go with them, and the I end up at a dead end...I appreciate all of your help!
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