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Hannah_Ahn

prove the identity: 1 over sec x + tan x = 1 - sin x over cos x

  • 2 years ago
  • 2 years ago

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  1. Hannah_Ahn
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    \[\left( 1 \over secx + tanx \right) = \left( 1-sinx \over \cos x \right)\]

    • 2 years ago
  2. Yifan12879
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    LHS=cosx / (1+sinx) LHS/RHS = (cos^2 x)/(1-sin^2 x)=cos^2 x / cos^2 x =1 so LHS=RHS

    • 2 years ago
  3. Hannah_Ahn
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    cosx / (1+sinx) ?

    • 2 years ago
  4. Yifan12879
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    yeah that is LHS

    • 2 years ago
  5. Yifan12879
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    multiply the numerator and denominator by cosx

    • 2 years ago
  6. Hannah_Ahn
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    oh yea

    • 2 years ago
  7. Hannah_Ahn
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    I don't get LHS/RHS? what's going on hehe

    • 2 years ago
  8. Yifan12879
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    divide LHS by RHS and found it is 1. so we can say LHS=RHS

    • 2 years ago
  9. Hannah_Ahn
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    (1-sinx) times (1+sinx) is one?

    • 2 years ago
  10. Yifan12879
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    (1-sinx) times (1+sinx) = |dw:1325741008297:dw|

    • 2 years ago
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