trig

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use a half-angle identity to find the exact value of tan 105 degrees.

  • trig -

    we know that
    tan 2A = 2tanA/(1 - tan^2 A)

    so tan 210 = 2tan105/(1 - tan^2 105)
    you should know that
    tan 210 = tan 30° = 1/√3

    let tan 105 = x

    then
    1/√3 = 2x/(1-x^2)
    2√3x = 1 - x^2
    x^2 + 2√3 - 1 = 0
    x = (-2√3 ± √16)/2
    = -√3 ± 2
    but obviously tan105 is negative, ( 105° is in II )
    so tan 105° = -√3-2

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