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Hannah_Ahn

  • 4 years ago

Prove the identity \[\left( \cot \theta \over \sin \theta - \csc \theta \right) = -\sec \theta\]

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  1. anonymous
    • 4 years ago
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    first we do the algebra \[\frac{\frac{a}{b}}{b-\frac{1}{b}}=\frac{a}{b^2-1}\]

  2. anonymous
    • 4 years ago
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    no replace a by cosine, b by sine and get \[\frac{\cos(x)}{\sin^2(x)-1}=-\frac{\cos(x)}{1-\sin^2(x)}=-\frac{\cos(x)}{\cos^2(x)}=-\sec(x)\]

  3. Hannah_Ahn
    • 4 years ago
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    \[a \over b^2 - 1\] ?????

  4. Hannah_Ahn
    • 4 years ago
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    oooh gotcha. i thought that was the right side,

  5. anonymous
    • 4 years ago
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    yes \[\frac{\frac{a}{b}}{b-\frac{1}{b}}=\frac{a}{b^2-1}\] multiply top and bottom by \[b\]and get it in one step

  6. Hannah_Ahn
    • 4 years ago
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    thanks! :)

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