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Math Help - Function Simplification

  1. #1
    Member rtblue's Avatar
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    Function Simplification

    Hey guys, this problem has been racking my brain for around an hour. I guess I'm just having a brain fart, as it should be easy. Either way, here it is:

    Simplify \displaysyle \frac{cot^2x}{cscx+1}+1

    This should simplify out to csc(x). Could anyone explain how that happens?

    Thanks in advance.
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  2. #2
    Member rtblue's Avatar
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    never mind guys... I found it.

    For those who want to know, just put everything under a common denominator. Then convert cot^2(x) + 1 to csc^2(x). Then factor, and it will cancel.

    Thanks guys!
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rtblue View Post
    Hey guys, this problem has been racking my brain for around an hour. I guess I'm just having a brain fart, as it should be easy. Either way, here it is:

    Simplify \displaysyle \frac{cot^2x}{cscx+1}+1

    This should simplify out to csc(x). Could anyone explain how that happens?

    Thanks in advance.
    I always start by replacing everything with sines and cosines:
    \displaystyle \frac{\frac{cos^2(x)}{sin^2(x)}}{\frac{1}{sin(x)} + 1} + 1

    \displaystyle \frac{\frac{cos^2(x)}{sin^2(x)}}{\frac{1}{sin(x)} + 1} \cdot \frac{sin^2(x)}{sin^2(x)} + 1

    \displaystyle \frac{cos^2(x)}{sin(x) + sin^2(x)} + 1

    Now add the 1 to the fraction and simplify.

    -Dan
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  4. #4
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    Hello, rtblue!

    It helps if you know this identity and its variations:

    . . \csc^2\theta \:=\:\cot^2\!\theta + 1


    \displaystyle \text{Simplify: }\;\frac{\cot^2\!x}{\csc x+1}+1

    \displaystyle \text{We have: }\:\frac{\cot^2\!x}{\csc x + 1} + 1

    . . . . . . \displaystyle =\;\frac{\csc^2\!x - 1}{\csc x + 1} + 1

    . . . . . . \displaystyle =\;\frac{(\csc x - 1)(\csc x + 1)}{\csc x + 1} + 1

    . . . . . . =\;(\csc x - 1) + 1

    . . . . . . =\; \csc x

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