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Homework Help: More trig

  1. May 28, 2007 #1
    more trig....

    Amm ive been at this problem for an hour already it loooked really easy but for some reason i cant reach the answer
    (tan x + sec x)^2 = (1 + sin x)/(1 - sin x)

    soo what went and tried was expand it and then exchage tan and sec..

    (sin^2 x /cos^2 x )+ (2/cos x) + (1/ sin^2 x)

    ((sin^2)(sin^2) + (2(sin^2)(cos)) + cos^2)/((sin^2)(cos^2))

    ((sin^2)(sin^2 + 2cos - 1) + 1)/((sin^2)(1-sin^2))

    (sin^2 + 2cos)/(1-sin^2)

    and after i get there i njust dont know how to get to the (1 + sin x)/(1 - sin x)

    any ideas??

  2. jcsd
  3. May 28, 2007 #2


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    you can write

    (1 + sin x)/(1 - sin x) = [(1 + sin x)(1 + sin x)]/[(1 + sin x)(1 - sin x)]
  4. May 28, 2007 #3


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    Oh, you're wrong from the start... =.="
    sec(x) = 1 / (cos(x)), instead of 1 / (sin(x)), as you have written above.
    You should re-do the problem, and follow malawi_glenn's hint.

    You can also manipulate both sides a the same time, instead of 1 side at 1 time. Like this:
    [tex](\tan x + \sec x) ^ 2 = \frac{1 + \sin x}{1 - \sin x}[/tex]
    [tex]\Leftrightarrow \left (\tan x + \frac{1}{\cos x} \right) ^ 2 = \frac{(1 + \sin x) ^ 2}{1 - \sin ^ 2 x}[/tex]
    <=> ...

    Can you go from here? :) It should be easy.
  5. May 29, 2007 #4
    ok well after asking arround i got to the answer... thanx for the help but the thing is im supposed to just use the properties to reach other side... i cant alter it... but i guess is my fault for not specifying... well the way it went was quite tricky and required a property i dont use normally which is tan^2x + 1 = sec^2 x

    so using that you get that

    2tan x sec x + sec^2x - 1... and one more thing i realised i was using the sec = 1/sin ... ummm opps... but really thanx for the ideas
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