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Babynini

  • one year ago

Prove the identity (tanx)/(1-cosx) = cscx(1+secx)

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  1. Babynini
    • one year ago
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    I've gotten up to (sinx/cosx)(1/(1-cosx)

  2. Babynini
    • one year ago
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    is that correct so far?

  3. freckles
    • one year ago
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    ok and you can write that as:\[\frac{\frac{\sin(x)}{\cos(x)}}{1-\cos(x)} \\ \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{1-\cos(x)}\]

  4. freckles
    • one year ago
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    oh that is what you wrote

  5. freckles
    • one year ago
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    lol

  6. freckles
    • one year ago
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    \[\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{1-\cos(x)} \cdot \frac{1+\cos(x)}{1+\cos(x)}\] see what this does maybe

  7. freckles
    • one year ago
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    multiply the second and last fraction

  8. Babynini
    • one year ago
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    so the last fraction is just like multiplying by 1 right?

  9. freckles
    • one year ago
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    yep

  10. Babynini
    • one year ago
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    It'd be (sinx/cosx)((1+cosx)/(1-cos^2x))

  11. freckles
    • one year ago
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    right and 1-cos^2(x) is sin^2(x)

  12. freckles
    • one year ago
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    \[\frac{\sin(x)}{\cos(x)} \cdot \frac{1+\cos(x)}{\sin^2(x)} \\ \frac{\sin(x)}{\sin^2(x)} \frac{1+\cos(x)}{\cos(x)}\]

  13. freckles
    • one year ago
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    can you see where to go from here?

  14. Babynini
    • one year ago
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    Emm no haha sorry.

  15. freckles
    • one year ago
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    \[\frac{\sin(x)}{\sin^2(x)}(\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)})\] how about here?

  16. freckles
    • one year ago
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    cos(x)/cos(x)=? 1/cos(x)=? sin(x)/sin^2(x)=?

  17. Babynini
    • one year ago
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    1) = 1 2) = secx 3) I'm not sure about that one :P

  18. Babynini
    • one year ago
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    where did you get the cosx/cosx?

  19. freckles
    • one year ago
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    1) good 2) good 3) do you know how to simplify say something like u/u^2 ? separated the fraction to "where did you get the cosx/cosx? " \[\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c} \\ \text{ we had } \frac{1+\cos(x)}{\cos(x)}=\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)}\]

  20. Babynini
    • one year ago
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    aah I see, ok :)

  21. Babynini
    • one year ago
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    3) um, i'm not sure.

  22. freckles
    • one year ago
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    \[\frac{u}{u^2}=\frac{u}{u \cdot u}=\frac{\cancel{u}}{\cancel{u} u}=\frac{1}{u} \\ \text{ or just use law of exponents } \\ \frac{u}{u^2}=u^{1-2}=u^{-1}=\frac{1}{u} \\ \frac{\sin(x)}{\sin^2(x)}=\frac{1}{\sin(x)}\]

  23. freckles
    • one year ago
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    and 1/sin(x) is....

  24. Babynini
    • one year ago
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    cscx!

  25. freckles
    • one year ago
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    yep

  26. freckles
    • one year ago
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    What we did: \[\frac{\tan(x)}{1-\cos(x)} \\ \tan(x) \cdot \frac{1}{1-\cos(x)} \\ \frac{\sin(x)}{\cos(x)} \frac{1}{1-\cos(x)} \\ \frac{\sin(x)}{\cos(x)} (\frac{1}{1-\cos(x) } \cdot 1) \\ \frac{\sin(x)}{\cos(x)} (\frac{1}{1-\cos(x) } \cdot \frac{1+\cos(x)}{1+\cos(x)} ) \\ \frac{\sin(x)}{\cos(x)}(\frac{1+\cos(x)}{1-\cos^2(x)}) \\ \frac{\sin(x)}{\cos(x)} \frac{1+\cos(x)}{\sin^2(x)} \\ \frac{\sin(x)}{\sin^2(x)} \frac{1+\cos(x)}{\cos(x)} \\ \frac{1}{\sin(x)} \frac{1+\cos(x)}{\cos(x)} \\ \frac{1}{\sin(x)} (\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)}) \\ \csc(x)(\sec(x)+1) \\ \csc(x)(1+\sec(x))\]

  27. Babynini
    • one year ago
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    aw thanks for taking the time to type that all up!

  28. Babynini
    • one year ago
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    haha I guess I should do a few more practice ones. Sigh. Thanks so much for the help :)

  29. freckles
    • one year ago
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    np trig identities can be tricky (Takes practice practice) but fun just think you are doing a puzzle or something puzzles should be fun

  30. Babynini
    • one year ago
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    Tricky may be an understatement xP haha sure sure.

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