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Homework Help: Math Problem

  1. Mar 2, 2006 #1
    Can anyone help me solve this equation,

     
  2. jcsd
  3. Mar 2, 2006 #2
    What have you done so far? Solutions are not just given away in this forum.

    Is it really "solve for x" or is it "prove this"?
     
  4. Mar 2, 2006 #3
    prove this.
    I can't seem to get rid of the bottom and make it 1 + tan(x) + cot(x)
     
    Last edited: Mar 2, 2006
  5. Mar 2, 2006 #4
    Hmmm, I haven't tried this problem, but just a tip, instead of writing cot as 1/tanx write it as cos/sin. Makes things simpler.

    The only thing I can suggest is get common denominators on the left side.

    Wait, are you trying to prove the identity?
     
  6. Mar 2, 2006 #5
    Hmm...
    [tex]
    \begin{gathered}
    \frac{{\cot x}}
    {{1 - \tan x}} + \frac{{\tan x}}
    {{1 - \cot x}} = 1 + \tan x + \cot x \Rightarrow \hfill \\
    \frac{1}
    {{\left( {1 - \tan x} \right)\tan x}} + \frac{{\tan ^2 x}}
    {{\tan x - 1}} = \frac{{\tan ^2 x + \tan x + 1}}
    {{\tan x}} \Rightarrow \hfill \\
    \frac{{1 - \tan ^3 x}}
    {{1 - \tan x}} = \frac{{\tan ^2 x + \tan x + 1}}
    {{\tan x}} \Rightarrow \hfill \\
    \left( {1 - \tan ^3 x} \right)\tan x = \left( {1 - \tan x} \right)\left( {\tan ^2 x + \tan x + 1} \right) \Rightarrow \hfill \\
    \tan x - \tan ^4 x = \tan ^2 x + \tan x + 1 - \tan ^3 x - \tan ^2 x - \tan x \Rightarrow \hfill \\
    \tan x\left( {1 - \tan ^3 x} \right) = 1 - \tan ^3 x \Rightarrow \tan x = 1 \Rightarrow {\text{Mistake?}} \hfill \\
    \end{gathered} [/tex]
     
  7. Mar 3, 2006 #6
    Line 3 should be
    [tex]\frac{1-tan^3x}{(1-tanx)tanx}=\frac{1+tanx+tan^2x}{tanx}[/tex]

    -Dan
     
  8. Mar 3, 2006 #7

    0rthodontist

    User Avatar
    Science Advisor

    Well, multiply through by tan x
    (1 - tan^3(x))/((tan x)(1 - tan x))
    Now you can factor the numerator as a difference of cubes, and reduce.
     
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