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AaronAndyson

  • one year ago

Perform the indicated expression: 2tan28°tan62°

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  1. ankit042
    • one year ago
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    x= 28 2tan(90-x).tan(x)= 2sin(90-x).sin(x)/cos(90-x).cos(x)

  2. ankit042
    • one year ago
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    Now sin(90-x)=? cos(90-x)=?

  3. Michele_Laino
    • one year ago
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    we have this: \[\Large \tan 90 = \tan \left( {28 + 62} \right) = \frac{{\tan 28 + \tan 62}}{{1 - \tan 28\tan 62}}\]

  4. Jhannybean
    • one year ago
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    Ahh, I didn't know you could combine it :)

  5. Michele_Laino
    • one year ago
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    now, since tan90= infinity, we have to request that: \[\Large 1 - \tan 28\tan 62 = 0\]

  6. Michele_Laino
    • one year ago
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    therefore: \[\Large 2 - 2\tan 28\tan 62 = 0\]

  7. AaronAndyson
    • one year ago
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    thanks

  8. AaronAndyson
    • one year ago
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    1/sec^2(36) - cot^(54) =

  9. AaronAndyson
    • one year ago
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    ^ help me with that

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