Trigonometry

Prove that
(tan4A + tan2A)(1-tan²3Atan²A) = 2tan3Asec²A

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  1. tan(3A+A) = (tan3A + tanA)/(1-tan3AtanA)
    tan(3A-A) = (tan3A - tanA)/(1+tan3AtanA)

    add them, and you have a common
    denominator of (1-tan²3Atan²A)
    numerator = (tan3A + tanA)(1+tan3AtanA) + (tan3A - tanA)(1-tan3AtanA)
    = 2(tan3A + tan3Atan²A)
    = 2tan3A(1+tan²A)
    = 2tan3Asec²2A

    1. 👍
    2. 👎
  2. tan(3A+A) = (tan3A + tanA)/(1-tan3AtanA)
    tan(3A-A) = (tan3A - tanA)/(1+tan3AtanA)

    add them, and you have a common
    denominator of (1-tan²3Atan²A)
    numerator = (tan3A + tanA)(1+tan3AtanA) + (tan3A - tanA)(1-tan3AtanA)
    = 2(tan3A + tan3Atan²A)
    = 2tan3A(1+tan²A)
    = 2tan3Asec²2A

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    2. 👎

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