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anonymous

  • one year ago

(sinx / one - cosx )+ (sin x /one - cosx) = 2 csc x

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  1. Loser66
    • one year ago
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    How??? check the problem, please.

  2. Loser66
    • one year ago
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    but it is true if the denominator is \(1-cos^2(x)\)

  3. anonymous
    • one year ago
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    what identity would i use @Concentrationalizing

  4. Loser66
    • one year ago
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    @Concentrationalizing please, give me the whole thing. Can't wait to see "the 2 more steps" to get the answer.

  5. anonymous
    • one year ago
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    Ah, I see what you mean. I wasn't paying attention, I apologize.

  6. anonymous
    • one year ago
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    so i would use the Pythagorean identity ? @Concentrationalizing

  7. anonymous
    • one year ago
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    No, I was wrong. Loser is right, the problem is faulty. I didnt even bother to check ahead, I just went through the motions of what steps you might normally do.

  8. anonymous
    • one year ago
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    so i'm confused what are the steps i have to do? @Concentrationalizing @Loser66

  9. anonymous
    • one year ago
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    The problem is incorrect. You maybe typed something wrong, but your question isn't true.

  10. anonymous
    • one year ago
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    As loser said, this is the set-up that would be correct http://www.wolframalpha.com/input/?i=sinx%2F%281-cos^2%28x%29%29+%2B+sinx%2F%281-cos^2%28x%29%29+%3D+2cscx+

  11. anonymous
    • one year ago
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    @Concentrationalizing

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  12. anonymous
    • one year ago
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    Okay, you typed it wrong in your original question then.

  13. anonymous
    • one year ago
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    \[\frac{ sinx }{ 1-cosx }+\frac{ sinx }{ 1+cosx } = \frac{ sinx(1+cosx)+sinx(1-cosx) }{(1+cosx)(1-cosx) }\] \[\frac{ sinx + sinxcosx + sinx - sinxcosx }{ 1-\cos^{2}x }\] \[\frac{ 2sinx }{ 1-\cos^{2}x }\] And yes, pythagorean identity from here. \(sin^{2}x + cos^{2}x = 1\) can be rearranged into \(sin^{2}x = 1-cos^{2}x\) Thus we have \[\frac{ 2sinx }{ \sin^{2}x }= \frac{ 2 }{ sinx } = 2cscx\]

  14. anonymous
    • one year ago
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    thank you @Concentrationalizing

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