Help with maths question: sin^2 x + tan^2 x + c

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    Hello, i need help on this question:


    sin2 x + tan2 x + cot2 x

    The answer is supposed to be equivalent to a single trig function, but using the above equation you're supposed to work out what it equals to.

    (Cot "cotangent" is Cos/sin)

    I've asked numerous peers and no one understands how to work it out, any help? I have NO clue on where to start, any help is appreciated.

    I've been told an example of what the answer should look like is something like Cosx, but that's an example to understand what your answer could possibly look like once simplifying it but it could look totally different.
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    sin^2(x) =1/2(1-Cos(2x))

    tan^2(x) = [1-cos(2x)]/[1+cos(2x)] (turn the tan into sin^2/cos^2 and use the identities for sin^2 = (1/2){1-cos^2()} etc.

    cot^2(x) =[ -1-cos(2x)]/-1+cos(2x)......
    EDIT: warning - but even then i can`t yet reduce it to a single function
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    (Original post by Hasufel)
    sin^2(x) =1/2(1-Cos(2x))

    tan^2(x) = [1-cos(2x)]/[1+cos(2x)] (turn the tan into sin^2/cos^2 and use the identities for sin^2 = (1/2){1-cos^2()} etc.

    cot^2(x) =[ -1-cos(2x)]/-1+cos(2x)......
    ok i've seen your edit, so what would be the simplest form of this equation?
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    (1/8)(26-Cos[2x]+6Cos[4x]+Cos[6x]) x (2Sin[2x]-1)
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    (Original post by Jakez123)
    ok i've seen your edit, so what would be the simplest form of this equation?
    I can only get this far,

     \displaystyle sin^2x + tan^2 x +cot^2x = \frac{1-cos^2x+cos^6x}{cos^2x(1-cos^2x)}
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    too knackered to try, but, just had a thought - in my post above, what about expanding the identities for Cos(ax)Sin(bx) and see if THAT simplifies (i`m going to bed now)!
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    (Original post by raheem94)
    I can only get this far,

     \displaystyle sin^2x + tan^2 x +cot^2x = \frac{1-cos^2x+cos^6x}{cos^2x(1-cos^2x)}
    Surely you can cancel the 1-cos^2 then you would get Cos^6x / cos^2x which cancels?
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    (Original post by shorttstuff)
    Surely you can cancel the 1-cos^2 then you would get Cos^6x / cos^2x which cancels?
    How?

    Can you show me a few steps?

    I don't think it is possible.

    And by the way, i didn't give you any neg rep, i don't know who gave you.
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    http://www.wolframalpha.com/input/?i...E2x%2Bcot%5E2x
    ......
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    I have tried wolframalpha but it doesn't gives the required answer.
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    Are you 100% sure "sin^2x + tan^2x + cot^2x" can be written as a single tri function?
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    Oh wait I dont think you can because of the 1-cos2x + cos6x

    And why the neg? (whoever it is) I only trying out a solution -.-
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    (Original post by RnTf)
    Are you 100% sure "sin^2x + tan^2x + cot^2x" can be written as a single tri function?
    well that's the thing to be honest, this question was on a test we had and we were told to simplify it, at most the smartest guy in the class only managed to simplify it to 2 trig functions and couldn't go further but people kept saying that you could simplify it to 1 trig function, so i won't lie i'm not even sure if you can do it, but I've stated in my third post that a sufficient answer would be the most simplistic answer you can get out of the equation.

    But everyone's posting different answers so now i'm getting even more confused. Sorry if his post wasn't much help.
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    (Original post by Jakez123)
    well that's the thing to be honest, this question was on a test we had and we were told to simplify it, at most the smartest guy in the class only managed to simplify it to 2 trig functions and couldn't go further but people kept saying that you could simplify it to 1 trig function, so i won't lie i'm not even sure if you can do it, but I've stated in my third post that a sufficient answer would be the most simplistic answer you can get out of the equation.

    But everyone's posting different answers so now i'm getting even more confused. Sorry if his post wasn't much help.
    \frac{1-cos^2x+cos^6x}{cos^2x(1-cos^2x)}

    This as was posted above, is simplified to (expressions of) a single trig function, which is what I'd believe the question meant, although it might have been worded unclearly. (Judging by the graph we can be sure you won't get it down to a single cosine or sine function)
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    (Original post by Jørgen)
    \frac{1-cos^2x+cos^6x}{cos^2x(1-cos^2x)}

    This as was posted above, is simplified to (expressions of) a single trig function, which is what I'd believe the question meant, although it might have been worded unclearly. (Judging by the graph we can be sure you won't get it down to a single cosine or sine function)
    Ok, thanks for telling me the answer, and i'm sorry to bother the community again, but i'm struggling to understand how you worked out the answer, forgive me for my ignorance but is it possible for a step-by-step guide to finding the answer you've shown? This is because if a similar question comes up i would need to work it out and most marks come from the working shown.

    Thanks in advance.
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    (Original post by Jakez123)
    Ok, thanks for telling me the answer, and i'm sorry to bother the community again, but i'm struggling to understand how you worked out the answer, forgive me for my ignorance but is it possible for a step-by-step guide to finding the answer you've shown? This is because if a similar question comes up i would need to work it out and most marks come from the working shown.

    Thanks in advance.
    Actually it was first worked out by me

     \displaystyle sin^2x +  tan^2x + cot^2x = (1 - cos^2x) + (sec^2x-1) + \left(\frac1{sec^2x-1}\right)

    Now write everything in terms of cosx and try to simplify it.
 
 
 
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