sec ⁡ θ + tan ⁡ θ = p and sec ⁡ θ tan ⁡ θ = q. Eliminate θ to form a equation between p and q. sec ⁡ θ + tan ⁡ θ = p ( sec ⁡ θ + tan ⁡ θ ) 2 = p 2 sec 2 ⁡ θ + tan 2 ⁡ θ + 2 tan ⁡ θ sec ⁡ θ = p 2 sec 2 ⁡ θ + tan 2 ⁡ θ + 2 q = p 2 1 + 2 tan 2 ⁡ θ + 2 q = p 2

Question

sec  ⁡  θ  +  tan  ⁡  θ  =  p and   sec  ⁡  θ  tan  ⁡  θ  =  q. Eliminate   θ to form a equation between p and q.  sec  ⁡  θ  +  tan  ⁡  θ  =  p  (  sec  ⁡  θ  +  tan  ⁡  θ      )    2    =      p    2        sec    2    ⁡  θ  +      tan    2    ⁡  θ  +  2  tan  ⁡  θ  sec  ⁡  θ  =      p    2        sec    2    ⁡  θ  +      tan    2    ⁡  θ  +  2  q  =      p    2    1  +  2      tan    2    ⁡  θ  +  2  q  =      p    2

assumintdz · Accepted Answer

Notice that      p    2    −  4  q  =  (  sec  ⁡  θ  −  tan  ⁡  θ      )    2      .Now       p    2    =  (  sec  ⁡  θ  +  tan  ⁡  θ      )    2  , so      p    2    (      p    2    −  4  q  )  =  (  sec  ⁡  θ  +  tan  ⁡  θ      )    2    (  sec  ⁡  θ  −  tan  ⁡  θ      )    2    =  (      sec    2    ⁡  θ  −      tan    2    ⁡  θ      )    2    =  1    .This probably seems a bit like magic. I actually first noticed that       p    2    −  2  q  =      sec    2    ⁡  θ  +      tan    2    ⁡  θ. Unfortunately, that didn’t seem to go anywhere. Then I noticed that subtracting another 2q would still give me something fairly nice, and the rest just fell into place.

Leland Morrow · Accepted Answer

Notice, we have                    (1)                    sec        ⁡        θ        +        tan        ⁡        θ        =        p                                (2)                    sec        ⁡        θ        tan        ⁡        θ        =        q              (  sec  ⁡  θ  −  tan  ⁡  θ      )    2    =  (  sec  ⁡  θ  +  tan  ⁡  θ      )    2    −  4  sec  ⁡  θ  tan  ⁡  θ  (  sec  ⁡  θ  −  tan  ⁡  θ      )    2    =      p    2    −  4  q                    (3)                    sec        ⁡        θ        −        tan        ⁡        θ        =                              p            2                    −          4          q                    adding (1) &amp;amp; (3),                    (4)                    2        sec        ⁡        θ        =        p        +                              p            2                    −          4          q                    &amp;amp; subtracting (3)from (1),                    (5)                    2        tan        ⁡        θ        =        p        −                              p            2                    −          4          q                    Now, squaring &amp;amp; subtracting (5) from (4), one should have  4      sec    2    ⁡  θ  −  4      tan    2    ⁡  θ  =            (      p      +                        p          2                −        4        q            )        2    −            (      p      −                        p          2                −        4        q            )        2    4  (      sec    2    ⁡  θ  −      tan    2    ⁡  θ  )  =      (    p    +                  p        2            −      4      q        +    p    −                  p        2            −      4      q        )        (    p    +                  p        2            −      4      q        −    p    +                  p        2            −      4      q        )    4  (  1  )  =      (    2    p    )        (    2                  p        2            −      4      q        )    p            p      2        −    4    q    =  1            p      2        (          p      2        −    4    q    )    =    1

sec ⁡ θ + tan ⁡ θ<!-

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