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Math Help - 2 questions: proving identity and reducing

  1. #1
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    2 questions: proving identity and reducing

    The first question I think I have the correct answer to but wanted to double check.

    The problem says prove the following identity...

    sec x + csc x = csc x + csc x tan x

    (1/cosx + 1/sin)=

    sinx/(sinxcosx) + cosx/(sinxcosx)


    (1/sinx)*(sinxcosx)= cscxtanx=secx

    Does this answer make sense?


    The second question I have I need to reduce this to a single term.

    tan (x+y) - tan y
    -----------------------
    1+ tan (x+y) tan y

    I know it reduces to tanx, but how would I show the work for a problem like this?

    Thanks so much for any help!
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  2. #2
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    Quote Originally Posted by KarlosK View Post
    The first question I think I have the correct answer to but wanted to double check.

    The problem says prove the following identity...

    sec x + csc x = csc x + csc x tan x

    (1/cosx + 1/sin)=

    sinx/(sinxcosx) + cosx/(sinxcosx)


    (1/sinx)*(sinxcosx)= cscxtanx=secx

    Does this answer make sense? no

    working from the right side ...

    \textcolor{red}{\csc{x} + \csc{x}\tan{x}}

    \textcolor{red}{\csc{x} + \frac{1}{\sin{x}} \cdot \frac{\sin{x}}{\cos{x}}}

    \textcolor{red}{\csc{x} + \frac{1}{\cos{x}}}

    \textcolor{red}{\csc{x} + \sec{x}}


    The second question I have I need to reduce this to a single term.

    tan (x+y) - tan y
    -----------------------
    1+ tan (x+y) tan y

    use this identity ...

    \textcolor{red}{\tan(a-b) = \frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}}

    let \textcolor{red}{a = (x+y)} and \textcolor{red}{b = y}
    ...
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  3. #3
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    Skeeter thanks for the help on the first problem, I now understand how you did that. On the second problem I am still a little unclear as to how I should show the work to the professor.

    So Tan A-B = x+y--y = x

    tan x.

    Is that enough?
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  4. #4
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    \frac{\tan(x+y) - \tan{y}}{1 + \tan(x+y)\tan{y}} = \tan[(x+y) - y] = \tan{x}
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