mmkoscielski
(__)^2= (csc x1)(csc x+1) solve for the blank



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mmkoscielski
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would it be cscx^2

BTaylor
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it would be \[\csc^{2} x\]

mmkoscielski
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okay

nolastudent
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(csc x1)(csc x+1) = \[\csc^{2}(x)  1 = (blank)^{2}\]
so \[\sqrt{\csc2(x)−1} = blank\]

gandalfwiz
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Nolastudent you're way makes sense, but is there a way to simplify is more?

nolastudent
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yes, sorry, trig identity: csc^2(x)1 = cot^2(x)

gandalfwiz
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*your*. Sorry... :)

nolastudent
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so square root of cotangent squared, which is cotangent.

nolastudent
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That's okay, we're in the math section XD

gandalfwiz
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Okay, that makes more sense. I can never simplify mine enough... :)

nolastudent
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LOL, thanks for the help gandalfwiz!

gandalfwiz
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So the final answer is cotx?

nolastudent
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yep