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Math Help - Help with verifying trig functions please!

  1. #1
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    Smile Help with verifying trig functions please!

    Hi, I'm having some trouble verifying a trig equation using identities. here is the problem:

    show that:

    (tanx-cotx)/(sinxcosx) = sec^2(x) - csc^2(x)

    i'm stuck, but I think the first thing I have to do is the pythagorean identity with tanx and cotx. is this right? if so what do i do next? (I'm working on left side of equation)
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  2. #2
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    Quote Originally Posted by TacticalPro View Post
    Hi, I'm having some trouble verifying a trig equation using identities. here is the problem:

    show that:

    (tanx-cotx)/(sinxcosx) = sec^2(x) - csc^2(x)

    i'm stuck, but I think the first thing I have to do is the pythagorean identity with tanx and cotx. is this right? if so what do i do next? (I'm working on left side of equation)
    \displaystyle\frac{\tan x-\cot x}{\sin x\cos x}=\frac{\tan x}{\sin x\cos x}-\frac{\cot x}{\sin x\cos x}=\frac{\frac{\sin x}{\cos x}}{\sin x\cos x}\cdots

    Get the idea?
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  3. #3
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    Quote Originally Posted by TacticalPro View Post
    Hi, I'm having some trouble verifying a trig equation using identities. here is the problem:

    show that:

    (tanx-cotx)/(sinxcosx) = sec^2(x) - csc^2(x)

    i'm stuck, but I think the first thing I have to do is the pythagorean identity with tanx and cotx. is this right? if so what do i do next? (I'm working on left side of equation)
    \frac{\tan x-\cot x}{\sin x * \cos x}


    =\frac{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}{\sin x * \cos x}


    =\frac{\frac{\sin ^2 x - \cos ^2 x}{\sin x * \cos x}}{\sin x * \cos x}



    =\frac{\sin ^2 x - \cos ^2 x}{\sin ^2 x * \cos^2 x}

    =\frac{\sin ^2 x}{\sin ^2 x * \cos^2 x}-\frac{\cos ^2 x}{\sin ^2 x * \cos^2 x}

    Simplify it and you will get the right hand side of your identity.
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  4. #4
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    Quote Originally Posted by Ithaka View Post
    \frac{\tan x-\cot x}{\sin x * \cos x}


    =\frac{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}{\sin x * \cos x}


    =\frac{\frac{\sin ^2 x - \cos ^2 x}{\sin x * \cos x}}{\sin x * \cos x}



    =\frac{\sin ^2 x - \cos ^2 x}{\sin ^2 x * \cos^2 x}

    =\frac{\sin ^2 x}{\sin ^2 x * \cos^2 x}-\frac{\cos ^2 x}{\sin ^2 x * \cos^2 x}

    Simplify it and you will get the right hand side of your identity.
    Thanks!

    I originally did that but got stuck at the end because i forgot I could split the fraction into two parts
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