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Yahoo!

  • 3 years ago

A+B = 5 pie/4, ( (1+cotA) ( 1 + cotB) )/(cotAcotB) = ?

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  1. hartnn
    • 3 years ago
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    ( (1+cotA) ( 1 + cotB) )/(cotAcotB) = (1+cot A + cot B + cot A cot B) / cotA cot B now separate the denominator.

  2. Yahoo!
    • 3 years ago
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    tanAtanB + TanA + tanB + 1

  3. hartnn
    • 3 years ago
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    what ?

  4. hartnn
    • 3 years ago
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    = (1+cot A + cot B)/cotAcot B + (cot A cot B) / cotA cot B = (1+cot A + cot B)/cotAcot B + 1

  5. hartnn
    • 3 years ago
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    that '1' will be a trouble...

  6. hartnn
    • 3 years ago
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    cot (A+B) =.... ?

  7. hartnn
    • 3 years ago
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    ok, lets work with tan tan(A+B) = tan 5pi/4 = ... ?

  8. Yahoo!
    • 3 years ago
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    tan 5pi/4=1

  9. hartnn
    • 3 years ago
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    tan(A+B) = (tan A+tan B) /(1+tan A. tan B) = ...

  10. Yahoo!
    • 3 years ago
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    tan(A+B) = (tan A+tan B) /(1-tan A. tan B)

  11. hartnn
    • 3 years ago
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    oops, tan(A+B) = (tan A+tan B) /(1-tan A. tan B) =1 which gives tan A+tan B = 1-tan A tan B

  12. hartnn
    • 3 years ago
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    so, tanAtanB + TanA + tanB + 1=...?

  13. hartnn
    • 3 years ago
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    tanAtanB + TanA + tanB + 1= tanAtanB +1-tan Atan B+ 1=2

  14. Yahoo!
    • 3 years ago
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    Yup... thxx...

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