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anonymous

  • one year ago

can somebody please help me with this question? i am completely lost. Use basic identities to simplify the expression. (picture below)

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  1. anonymous
    • one year ago
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  2. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{ \cos^2\theta}{\sin^2\theta}+\csc\theta\sin\theta }\)

  3. SolomonZelman
    • one year ago
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    some basic identities: \(\large \color{black}{\sin\theta\csc\theta=1}\) \(\large \color{black}{\cos\theta\sec\theta=1}\) \(\large \color{black}{\tan\theta\cot\theta=1}\)

  4. SolomonZelman
    • one year ago
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    i still want to ask, do you have any options?

  5. anonymous
    • one year ago
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    these are my options

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  6. SolomonZelman
    • one year ago
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    ok, then you need 2 more rules" \(\color{black}{\large \dfrac{\sin\theta}{\cos\theta}=\tan\theta}\) \(\large \color{black}{\tan^2\theta+1=\sec^2\theta}\)

  7. SolomonZelman
    • one year ago
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    well, the first rule also applies when sine and cosine are raised to any power of n, such that: \(\color{black}{\large \dfrac{\sin^n\theta}{\cos^n\theta}=\tan^n\theta}\) (for all n, of course)

  8. Nnesha
    • one year ago
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    or maybe u need 2 More rules :3 Mo\[\huge\rm \frac{ \cos \theta }{ \sin \theta}= \cot \theta \] \[\huge\rm cot^2 \theta +1=\csc^2\theta \]

  9. SolomonZelman
    • one year ago
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    oh, Nnesha, you are right, because it is cosine/sine, not the other way around as I thought for some reason:) and thus, the 2nd rule you posted comes about.

  10. anonymous
    • one year ago
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    so, what exactly do i do with the rules you guys gave me?

  11. SolomonZelman
    • one year ago
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    \(\LARGE \color{blue}{\unicode{x10009999999}}\)\(\LARGE \color{blue}{\unicode{x120D999}}\)\(\LARGE \color{blue}{\unicode{x3D5}}\)\(\LARGE \color{blue}{\unicode{x8594}}\) \(\LARGE \color{blue}{:)}\)

  12. SolomonZelman
    • one year ago
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    u apply them to your problem

  13. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{ \cos^2\theta}{\sin^2\theta}+\csc\theta\sin\theta }\) apply to the second term, one of the rules that I posted in the very beginning.

  14. anonymous
    • one year ago
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    okay what i think i understand is cos^2(theta) / sin^2(theta) = 1

  15. anonymous
    • one year ago
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    then cos/sin = cot

  16. anonymous
    • one year ago
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    would it then be cot^2(theta) = 1 ?

  17. Nnesha
    • one year ago
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    nope.

  18. Nnesha
    • one year ago
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    cos^2/sin^2 is not equal to one if cos /sin =cot then cos^2 / sin^2 = cot^2

  19. anonymous
    • one year ago
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    cot^2 (theta) + 1 ?

  20. Nnesha
    • one year ago
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    yes right cot^2 theta +1 which is e qual to what ?

  21. anonymous
    • one year ago
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    csc^2 (theta) ?

  22. Nnesha
    • one year ago
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    yep right

  23. anonymous
    • one year ago
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    thank you so much!

  24. Nnesha
    • one year ago
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    my pleasure

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