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Math Help - False identity?

  1. #1
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    False identity?

    Hello, all!

    I need to prove the following identity:

    \frac{cos(x-y)}{sin(x-y)} = \frac{1 + tan\hspace{5 pt}x\hspace{5 pt}tan\hspace{5 pt}y}{tan\hspace{5 pt}x + tan\hspace{5 pt}y}

    I am able to prove that the left side of the equation is equal to:

    \frac{1 + tan\hspace{5 pt}x\hspace{5 pt}tan\hspace{5 pt}y}{tan\hspace{5 pt} x - tan\hspace{5 pt}y}

    The obvious difference between what I am able to prove and what the identity actually says is the difference in sign between tan x and tan y in the denominator.

    I'll be happy to explain how I arrived at my proof, but I'm hoping it should be pretty straight forward to understand.

    So my question is this:

    Is there some way to prove this identity, or is this a false identity (perhaps a misprint in the text I'm reading)?

    Thank you!
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  2. #2
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    How did you do it? I would expand the LHS.
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  3. #3
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    Yes, I expanded the LHS using addition laws. I then multiplied both the numerator and denominator by:

    \frac{1}{cos\hspace{5pt}x\hspace{5pt}cos\hspace{5p  t}y}

    That yields the answer I provided, with tan x - tan y in the denominator.
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  4. #4
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    Hello, JennyFlowers!

    I need to prove the following identity: . \frac{\cos(x-y)}{\sin(x-y)} \;=\; \frac{1 + \tan x \tan y}{\tan x + \tan y} . Not true!

    I am able to prove that the left side of the equation is equal to: . \frac{1 + \tan x \tan y}{\tan x -\tan y} . You are right!

    Is there some way to prove this identity, or perhaps a misprint in the text?
    . . It's a misprint.

    \frac{\cos(x-y)}{\sin(x-y)} \;=\;\cot(x-y) \;=\;\frac{1}{\tan(x-y)} \;=\;\frac{1}{\dfrac{\tan x - \tan y}{1 + \tan x\tan y}} \;=\;\frac{1 + \tan x\tan y}{\tan x - \tan y}

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  5. #5
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    Thank you!
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