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Math Help - Trigonometry Question

  1. #1
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    Angry Trigonometry Question

    Please Help Me in proving this----- If 3secA-7tanA=3, prove that 3tanA-7secA=7
    Last edited by raj1985; February 10th 2009 at 10:16 PM. Reason: Not complete
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  2. #2
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    Hello, raj1985!

    Is there a typo? .The statement isn't true.


    Prove: .If 3\sec A-7\tan A\:=\:3, then: . 3\tan A-7\sec A\:=\:{\color{red}-}7 . ??

    We have: . 3\sec A \:=\:3  + 7\tan A

    Square both sides: . 9\sec^2\!A \:=\:9 + 42\tan A + 49\tan^2\!A

    . . . . . . . . . . . 9(\tan^2\!A + 1) \:=\:9 + 42\tan A + 49\tan^2\!A

    . . . . . . . . . . . 9\tan^2\!A + 9 \:=\:9 + 42\tan A + 49\tan^2\!A

    . . . . . . 40\tan^2\!A - 42\tan A \:=\:0

    . . . . . . 2\tan\theta(20\tan A - 21) \:=\:0


    We have: . 2\tan\theta \:=\:0\quad\Rightarrow\quad \theta \,=\,0 .
    for which the statement is true.


    And: . 20\tan A - 21 \:=\:0\quad\Rightarrow\quad \tan A \:=\:\frac{21}{20}

    . . . \sec^2\!A \:=\:\tan^2\!A + 1 \:=\:\left(\frac{21}{20}\right)^2 + 1 \:=\:\frac{841}{400} \quad\Rightarrow\quad \sec A \:=\:\frac{29}{20}

    Then: . 3\tan A - 7\sec A \;=\;3\left(\frac{21}{20}\right) - 7\left(\frac{29}{20}\right) \;=\;\frac{63}{20} - \frac{203}{20} \;=\;\frac{\text{-}140}{20} \;=\;{\color{red}-7}

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