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March 26, 2017

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Prove the following:
1/(tanØ - secØ ) + 1/(tanØ + secØ) = -2tanØ

(1 - sinØ)/(1 + sinØ) = sec^2Ø - 2secØtanØ + tan^2Ø

  • Trig - ,

    That's a better job of typing it.
    I did the first of these in your previous post..

    the 2nd:

    LS = (1-sinØ)/(1+ sinØ)
    = (1-sinØ)/(1+ sinØ) * (1-sinØ)/(1- sinØ)
    = (1 - 2sinØ + sin^2 Ø)/(1 - sin^2 Ø)
    = (1 - 2sinØ + sin^2 Ø)/cos^2 Ø
    = 1/cos^2 Ø - 2sinØ/cos^2 Ø + sin^2 Ø/cos^2Ø
    = sec^2 Ø - 2(sinØ/cosØ)*(1/cosØ) + tan^2 Ø
    = sec^2 Ø - 2tanØ secØ + tan^2 Ø
    = RS

  • Trig - ,

    or, if you divide top and bottom by cosØ you have

    (secØ-tanØ)/(secØ+tanØ)

    now multiply top and bottom by (secØ-tanØ) and you have

    (secØ-tanØ)^2 / (sec^2Ø-tan^2Ø)
    = sec^2Ø - 2secØtanØ + tan^2Ø

  • Trig - ,

    Thanks to the both of you :)

  • Trig - ,

    hj

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