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Math Help - Trigonometric Proof

  1. #1
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    Trigonometric Proof

    Hi there,

    I need help with proving two trigonometric expressions:

    1) tanA - tanB = tanB tan(A-B)
    tanA - cotB


    2) tan (x - y) + tany
    tan (x - y) - coty

    Thanks in advance!
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  2. #2
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    Can anyone show me how to prove these questions?
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  3. #3
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    hello friend, as far as i know, trig. expressions are not proven, they are usually simplified....are the above two questions identities that you want help with?

    and with the second, if its proving, what do you want us to prove?

    and for the first, do you want us to prove this:

    \frac{\tan A - \tan B} {\tan A - \cot B } = \tan B \tan (A-B)

    - sugarT
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  4. #4
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    Quote Originally Posted by Sallyvinka View Post
    Hi there,

    I need help with proving two trigonometric expressions:

    1) tanA - tanB = tanB tan(A-B)
    tanA - cotB


    2) tan (x - y) + tany
    tan (x - y) - coty

    Thanks in advance!
    In the first problem there is a typo. It should be

    \frac{tanA - tanB}{tanA + cotB}

    Put cot(B) = 1/tan(B) simplify.

    In the second problem put cot(y) = 1/tan(y). The problem becomes

    \frac{tan(x-y) + tan(y)}{tan(x-y) - \frac{1}{tan(y)}}

    Now simplify.
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  5. #5
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    The question in my textbook says prove that you can get from the left hand side to the right hand side using trigonometric equations for:

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  6. #6
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    Quote Originally Posted by Sallyvinka View Post
    The question in my textbook says prove that you can get from the left hand side to the right hand side using trigonometric equations for:

    tan(A - B) = [( tan(A) - tan(B)]/[1 + tan(A)tan(B)]

    So there must a print mistake in the problem.
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  7. #7
    Newbie sugarT's Avatar
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    ye there must, i tried working it out, it don't lead to what your textbook was asking
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  8. #8
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    Hello Sallyvinka
    Quote Originally Posted by Sallyvinka View Post
    Hi there,

    I need help with proving two trigonometric expressions:

    1) tanA - tanB = tanB tan(A-B)
    tanA - cotB
    As others have pointed out, there is a mistake in this question. It should read:
    \frac{\tan A + \tan B}{\tan A - \cot B}=\tan B \tan(A-B)
    2) tan (x - y) + tany
    tan (x - y) - coty
    You don't say what you want to prove here. But perhaps it's this:
    \frac{\tan(x-y)+\tan y}{\tan(x-y) -\cot y}
    =\frac{\tan y(\tan(x-y)+\tan y)}{\tan y\tan(x-y) -1}, multiplying 'top-and-bottom' by \tan y

    =-\tan y\cdot\frac{\tan(x-y)+\tan y}{1-\tan y\tan(x-y)}, re-arranging


    =-\tan y \cdot\tan([x-y]+y), using \tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}

    =-\tan x \tan y
    Grandad
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  9. #9
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    I solved the question, thanks for everyone's help!
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