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ballerinaxbb

  • 2 years ago

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  1. ballerinaxbb
    • 2 years ago
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    |dw:1352855287144:dw|

  2. Schrodinger
    • 2 years ago
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    is that:\[\frac{ \cos(x) }{ 1 + \sin(x) }+ \frac{ 1 + \sin(x) }{ \cos(x) } = 2\sec(x)\]? (Just making sure.)

  3. ballerinaxbb
    • 2 years ago
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    Yes, that's exactly right

  4. Schrodinger
    • 2 years ago
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    Okay. The way I was taught in general to solve problems like this is always to take the more complex side, which was generally the one with two terms, and then work towards the side with less. That being said, (one second.)

  5. Schrodinger
    • 2 years ago
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    I guess what you could first do it multiply both of your numerators and denominators by the opposite denominator to make sure both numerators can be dealt with. e.g.: \[\frac{ \cos(x) }{1+\sin(x) }+\frac{ 1+\sin(x) }{\cos(x) }=2\sec(x)\]...\[\frac{ (cosx)(cosx) + (1+sinx)(1+sinx) }{(1+sinx)(cosx)}\]...\[\frac{ \cos ^{2}x + \sin ^{2}x +2sinx+1 }{ sinxcosx + cosx }\]From here, \[\sin ^{2}x+\cos ^{2}x\]can be simplified to 1. Follow so far?

  6. ballerinaxbb
    • 2 years ago
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    Ok... I think I get it What comes next?

  7. Schrodinger
    • 2 years ago
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    Sorry, multitasking. Well, let's make sure you do. Is there anything, ANYTHING at all you don't understand? I'm not your math teacher, and you're not in a classroom, don't be afraid to ask if you don't get something.

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