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Help with proofs

  1. Apr 13, 2005 #1
    I need help solving 2 proofs:

    tan x + cot x = (sec x)(csc x)

    I changed the left side to:

    tan x + 1/tan x = (sec x)(csc x)

    then crossed out the tan:

    1 = (sec x)(csc x), but I got stuck there.

    The next one I had trouble with was:

    tan^2 x - sin^2 x = (tan^2 x)(sin^2 x)

    I saw the left side being a^2 - b^2, so I factored it into:

    (tan x + sin x)(tan x - sin x) = (tan^2 x)(sin^2 x)

    I then changed the tan into sin/cos:

    ((sin x/cos x) + sin x)) ((sin x/cos x) - sin x)) , but got stuck there.

    Can you help me solve these proofs?
     
  2. jcsd
  3. Apr 13, 2005 #2

    dextercioby

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    Try to write everything in terms of sine & cosine...You'll get them easily.

    Daniel.
     
  4. Apr 13, 2005 #3
    So for the second one:

    tan^2 x - sin^2 x = (tan^2 x)(sin^2 x)

    (sin^2 x/cos^2 x) - sin^2 X = (sin^2 x/cos^2 x)(sin^2x)

    So on both sides so the sin^2 x cancel, leaving it like:

    cos^2 x = cos^2 x?

    And for the first one:

    tan x + cot x = (sec x)(csc x)

    I changed it to:

    sin x/cos x + 1/(sin x/cos x) = (1/cos x)(1/sin x)

    What would I do from here?
     
  5. Apr 13, 2005 #4

    dextercioby

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    Bring it to the same denominator (in the LHS) and after simplifying the denominators,u'll find

    [tex] \sin^{2}x+\cos^{2}x =1 [/tex]



    Daniel.
     
  6. Apr 14, 2005 #5

    HallsofIvy

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    "crossed out the tan" is not a mathematics term! I'm serious- think about exactly what you are doing there. tan x+ 1/tan x is NOT equal to 1 for all x!

    Okay, sin^2 x/cos^2 x) - sin^2 x= (sin^2 x)((1/cos^2x) - 1) so canceling sin^2 x leaves (1/cos^2 x)- 1 = sin^2 x/cos^2 x

    That is NOT "cos^2 x= cos^2 x" but if you multiply both sides by cos^2 x you get something almost as easy.
     
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