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Math Help - trig identities!!!help!!!

  1. #1
    awesometrackgirl
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    trig identities!!!help!!!

    i need major help with some problems, i want to be able to figure them out on my own, so please don't give me the answers

    okay i got so far, but i don't think i did it right
    here's the problem

    (cosx)(sec3x+cotx)=tan2x\1-cos2x+sinx\sec2-1

    (sinx\cosx)(sinx\cosx)/sin2x+sin\tan

    then i crossed out the sins but i don't think i did it right, please help!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by awesometrackgirl
    i need major help with some problems, i want to be able to figure them out on my own, so please don't give me the answers

    okay i got so far, but i don't think i did it right
    here's the problem

    (cosx)(sec3x+cotx)=tan2x\1-cos2x+sinx\sec2-1

    (sinx\cosx)(sinx\cosx)/sin2x+sin\tan

    then i crossed out the sins but i don't think i did it right, please help!!
    It looks like you are asking for help in proving the following trig identity:

    cos(x)[sec^3(x)+cot(x)]=tan^2(x)/(1-cos^2(x)) + sin(x)/(sec^2(x)-1)

    which is how we would normaly write this is plain ASCII.

    It is easier to see what this is if we rewrite it in TeX:

    <br />
\cos(x)[\sec^3(x)+\cot(x)]=\frac{\tan^2(x)}{1-\cos^2(x)} + \frac{\sin(x)}{\sec^2(x)-1}<br />

    RonL
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  3. #3
    Grand Panjandrum
    Joined
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    We wish to prove the trig identity:

    <br />
\cos(x)[\sec^3(x)+\cot(x)]=\frac{\tan^2(x)}{1-\cos^2(x)} + \frac{\sin(x)}{\sec^2(x)-1}<br />

    Focus on the RHS, look at the first term:

    <br />
\frac{\tan^2 (x)}{1-\cos^2(x)} <br />

    use \sin^2(x)=1-cos^2(x) to simplify the denominator, then try to
    rearrange what you have to be the first term on the LHS after the bracket is
    expanded, i.e. the first term of:

    <br />
\cos(x)\sec^3(x)+\cos(x)\cot(x)<br />

    Now repeat the process with the second term on the RHS (and show it is
    equal to the second term in the expression above.

    RonL
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