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Hannah_Ahn

  • 4 years ago

prove the identity: \[\left( \cot x - 1 \over 1 - \tan x\right) = \left( \csc x \over \sec x \right)\]

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  1. anonymous
    • 4 years ago
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    left hand side is \[\frac{\frac{1}{b}}{\frac{1}{a}}=\frac{a}{b}\]

  2. anonymous
    • 4 years ago
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    sorry that was right hand side

  3. Hannah_Ahn
    • 4 years ago
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    it's okay (-:

  4. anonymous
    • 4 years ago
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    left hand side is \[\frac{\frac{a}{b}-1}{1-\frac{b}{a}}=\frac{a^2-ab}{ab-b^2}=\frac{a(a-b)}{b(a-b)}=\frac{a}{b}\] so there are the same. there is no trig here

  5. Hannah_Ahn
    • 4 years ago
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    that is an awesome way to prove it! you are amazing! thanks!!!

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