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bridgetx516x

  • 2 years ago

Simplify the expression as much as possible: cscx / tanx+cotx

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  1. paddyfitz
    • 2 years ago
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    \[\frac{ cosec(x) }{ \tan(x)+\cot(x) }= \frac{ \frac{ 1 }{ \sin(x) } }{ \frac{ \sin(x) }{ \cos(x) }+\frac{ \cos(x) }{ \sin(x) } }\] So if we move the sin(x) to the bottom like so \[= \frac{ 1 }{ \sin(x)\frac{ \sin(x) }{ \cos(x) }+ \sin(x)\frac{ \cos(x) }{ \sin(x) } }\] Expand out the bottom further to get \[= \frac{ 1 }{ \frac{ \sin^2(x) }{ \cos(x) }+\cos(x) }\] Now let cos(x)=cos^2(x)/cos(x) and take out a factor of 1/cos(x) on the bottom.\[=\frac{ 1 }{ \frac{ \sin^2(x)+\cos^2(x) }{ \cos(x) } }\]As we know, sin^2(x) + cos^2(x) = 1, so we are left with 1/(1/cos(x)) which is just \[\frac{ cosec(x) }{ \tan(x)+\cot(x) }= \cos(x)\]Hope this helped:)

  2. NoelGreco
    • 2 years ago
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    Paddy, you should have gotten a medal or at least a thank you for all that work!

  3. NoelGreco
    • 2 years ago
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    There, I gave you a medal.

  4. bridgetx516x
    • 2 years ago
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    THANK YOU SO MUCH!

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