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Precalculus HW problem

  1. Oct 25, 2006 #1
    Can anyone help me sove this problem?

    cot x + tan x + 1 = (cot x / 1 - tan x) + (tan x / 1 - cot x)
     
    Last edited: Oct 25, 2006
  2. jcsd
  3. Oct 25, 2006 #2
    convert the right hand sides to sines and cosines, add the fractions, and simplify. or you could use the fact that tan x = 1 / cot x
     
    Last edited: Oct 25, 2006
  4. Oct 25, 2006 #3
    Since right hand side is more complex, I think we should start from that side, then arrive at the left hand side.

    We can use the properties of that cot x is the reciprocal of tan x, too.
     
  5. Oct 25, 2006 #4
    Here is what i got by working on the right hand side.

    (cot x / 1- (sin x / cos x)) + (tan x / 1 - (cos x / sinx)) = (cot x / (cos x - sin x / cos x)) + (tan x / (sin x - cos x / sin x))

    then i tried several different ways to work on the problem from this step, but it never worked.
     
  6. Oct 25, 2006 #5
    convert the numerators to sines and cosines as well.
     
  7. Oct 25, 2006 #6
    here is what i got

    (cos x / sin x ) / (cos x - sin x / cos x ) + (sin x / cos x) / (sin x - cos x / sin x)
    = (cos x / sin x )*(cos x / cos x - sin x) + (sin x / cos x)*(sin x / sin x - cos x)
    = (cos ^2 x / cos x - sin^2 x) + (sin^2 x / sin x - cos^2 x)
    = (sin^2 x - 1 / cos x - sin^2 x ) + (cos^2 x - 1 / sin x - cos^2 x)
    I tried to work on the problem until i got to this step, but I'm not sure if I'm on the right track.
     
  8. Oct 25, 2006 #7
    use the identity tan x = 1 / cot x instead.

    So [tex] \cot x + \tan x + 1 = \frac{\cot x}{1- \frac{1}{\cot x}} [/tex] etc..
     
    Last edited: Oct 25, 2006
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