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Math Help - I having a headache proving an identity

  1. #1
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    I having a headache proving an identity

    I having some problem proving a identity.

    Prove:
     <br />
(sec^2x )/tanx = (cot^2x - tan^2x) / (cotx - tanx)<br />




    i start by the left side so i get (1+tan^2x) / tan
    , and i then i convert tans to sins and cons, so i get (1+(sinX^2/con^2x))/(sinx/sin)

    after that i am lost, i dont know if i should start by the left side.

    plz help
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  2. #2
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    Quote Originally Posted by darkmoon123 View Post
    I having some problem proving a identity.

    Prove:
     <br />
(sec^2x )/tanx = (cot^2x - tan^2x) / (cotx - tanx)<br />




    i start by the left side so i get (1+tan^2x) / tan
    , and i then i convert tans to sins and cons, so i get (1+(sinX^2/con^2x))/(sinx/sin)

    after that i am lost, i dont know if i should start by the left side.

    plz help
    I would start with the left hand side b/c it is the difference of squares

    \frac{(\cot(x)+tan(x))(\cot(x)-\tan(x))}{\cot(x)-tan(x)}=\cot(x)+tan(x)=\frac{\cos(x)}{\sin(x)}+\fr  ac{\sin(x)}{\cos(x)}

    =\frac{\cos^{2}(x)+\sin^2(x)}{\sin(x)\cos(x)}=\fra  c{1}{\sin(x)\cos(x)}\cdot \left( \frac{\frac{1}{\cos(x)}}{\frac{1}{\cos(x)}}\right)  =\frac{\sec^2(x)}{\tan(x)}
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  3. #3
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    Hello, darkmoon123!

    Prove: .  \frac{\sec^2x}{\tan x} \:=\:\frac{\cot^2\!x - \tan^2\!x}{\cot x - \tan x}
    I would start with the right side . . .


    \frac{\cot^2\!x-\tan^2\!x}{\cot x -\tan x} \;=\;\frac{(\cot x-\tan x)(\cot x + \tan x)}{\cot t - \tan x} \;=\;\cot x + \tan x

    . . = \;\frac{1}{\tan x} + \tan x \;=\;\frac{\overbrace{1+\tan^2\!x}^{\text{This is }\sec^2\!x}}{\tan x} \;=\;\frac{\sec^2\!x}{\tan x}

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  4. #4
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    I actually wanted to resolve the hardest part which is is the left compressed part, so i need to expand until it looks like the right part.


     <br />
(1+tan^2x) / (tanx)<br />


     <br />
= ((cos^2x ) / (cos^2x) + (sin^2x) / (cos^2x)) / (sinx)/(cos)<br />


    but it fails me.
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  5. #5
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    Quote Originally Posted by darkmoon123 View Post



    I actually wanted to resolve the hardest part which is is the left compressed part, so i need to expand until it looks like the right part.


     <br />
(1+tan^2x) / (tanx)<br />


     <br />
= ((cos^2x ) / (cos^2x) + (sin^2x) / (cos^2x)) / (sinx)/(cos)<br />


    but it fails me.
    \frac{1+tan^2(x)}{tan(x)} = \frac{1}{tan(x)} + \frac{tan^2(x)}{tan(x)} = cot(x) + tan(x)

    Multiply by \frac{cot(x)-tan(x)}{cot(x)-tan(x)} which will give the same denominator as the right hand side.

    \frac{(cot(x)+tan(x))(cot(x)-tan(x))}{cot(x)-tan(x)}

    noting that the numerator is the difference of two squares we get the RHS

    \frac{(cot^2x - tan^2x)}{(cotx - tanx)}

    I can't take the credit for this, it's essentially what TheEmptySet did in reverse
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