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AaronAndyson

  • one year ago

(1+tanA.tanB)^2 + (tanA - tanB)^2 = sec(A)^2 sec(B)^2

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  1. AaronAndyson
    • one year ago
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    @welshfella

  2. AaronAndyson
    • one year ago
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    @Michele_Laino

  3. Michele_Laino
    • one year ago
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    we can try to use these identities: \[\begin{gathered} \tan A = \frac{{\sin A}}{{\cos A}} \hfill \\ \hfill \\ \tan B = \frac{{\sin B}}{{\cos B}} \hfill \\ \end{gathered} \]

  4. AaronAndyson
    • one year ago
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    ????

  5. Michele_Laino
    • one year ago
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    we can rewrite left side as follows:

  6. AaronAndyson
    • one year ago
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    as???

  7. Michele_Laino
    • one year ago
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    \[\begin{gathered} {\left( {1 + \tan A\tan B} \right)^2} = {\left( {1 + \frac{{\sin A}}{{\cos A}}\frac{{\sin B}}{{\cos B}}} \right)^2} = \hfill \\ \hfill \\ = {\left( {\frac{{\cos A\cos B + \sin A\sin B}}{{\cos A\cos B}}} \right)^2} = \hfill \\ \hfill \\ = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} \hfill \\ \end{gathered} \]

  8. Michele_Laino
    • one year ago
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    it the first term

  9. Michele_Laino
    • one year ago
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    the second term will be as follows:

  10. AaronAndyson
    • one year ago
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    ?

  11. Michele_Laino
    • one year ago
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    \[\begin{gathered} \tan A - \tan B = \frac{{\sin A}}{{\cos A}} - \frac{{\sin B}}{{\cos B}} = \hfill \\ \hfill \\ = \frac{{\sin A\cos B - \cos A\sin B}}{{\cos A\cos B}} = \hfill \\ \hfill \\ = \frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}} \hfill \\ \hfill \\ \end{gathered} \]

  12. AaronAndyson
    • one year ago
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    ok!

  13. Michele_Laino
    • one year ago
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    I think that maybe there is a typo in your original equation

  14. Michele_Laino
    • one year ago
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    please check your identity

  15. AaronAndyson
    • one year ago
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    Yep.Fixed it.

  16. Michele_Laino
    • one year ago
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    ok! so we can continue, and we can write this:

  17. Michele_Laino
    • one year ago
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    \[{\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2}\]

  18. AaronAndyson
    • one year ago
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    ok!

  19. Michele_Laino
    • one year ago
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    then we can write this next step: \[\begin{gathered} {\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2} = \hfill \\ \hfill \\ = \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} + \frac{{{{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\ \end{gathered} \]

  20. AaronAndyson
    • one year ago
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    ok!

  21. Michele_Laino
    • one year ago
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    therefore, we get: \[\begin{gathered} {\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2} = \hfill \\ \hfill \\ = \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} + \frac{{{{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2} + {{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{1}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\ \end{gathered} \]

  22. AaronAndyson
    • one year ago
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    may you explain the second last step?

  23. Michele_Laino
    • one year ago
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    in general we have this identity: (cos x)^2+ (sin x)^2 =1 for every value of x

  24. AaronAndyson
    • one year ago
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    ok

  25. Michele_Laino
    • one year ago
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    now we use these identities: \[\begin{gathered} \sec A = \frac{1}{{\cos A}} \hfill \\ \hfill \\ \sec B = \frac{1}{{\cos B}} \hfill \\ \hfill \\ \end{gathered} \]

  26. AaronAndyson
    • one year ago
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    ok

  27. Michele_Laino
    • one year ago
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    so we can rewrite the right side as follows: \[\begin{gathered} {\left( {\sec A} \right)^2}{\left( {\sec B} \right)^2} = {\left( {\frac{1}{{\cos A}}} \right)^2}{\left( {\frac{1}{{\cos B}}} \right)^2} = \hfill \\ \hfill \\ = \frac{1}{{{{\left( {\cos A} \right)}^2}}}\frac{1}{{{{\left( {\cos B} \right)}^2}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered} \]

  28. AaronAndyson
    • one year ago
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    thanks

  29. Michele_Laino
    • one year ago
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    :)

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