anonymous
  • anonymous
what is (cosx/sinx)/sinxcosx simplified
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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abb0t
  • abb0t
\[\frac{ \cos(x) }{ \sin(x) } \times \frac{ \sin(x)\cos(x) }{ 1 }\]
zepdrix
  • zepdrix
\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x}\]Woops I dunno if you're looking at it correctly abbot :O Only the top one is a fraction.
zepdrix
  • zepdrix
\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x} \qquad = \qquad \left(\frac{\cos x}{\sin x}\right)\frac{1}{\sin x \cos x}\]

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anonymous
  • anonymous
ya but what about multiplying by the recipricol.... cause i am seriously confused.
zepdrix
  • zepdrix
\[\large \left(\frac{\cancel{\cos x}}{\sin x}\right)\frac{1}{\sin x \cancel{\cos x}}\]
zepdrix
  • zepdrix
Ok here's the thing with the reciprocal :) lemme explain.
anonymous
  • anonymous
ok sounds good
zepdrix
  • zepdrix
The bottom fraction is actually this,\[\large \frac{\sin x \cos x}{1}\] So if you wanted to write it as a division of fractions, you could write it like this,\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x} \qquad = \qquad \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\left(\dfrac{\sin x \cos x}{1}\right)}\]And then from here, we could multiply by the reciprocal of the bottom fraction.
anonymous
  • anonymous
aha!!!!! now it's making sense!!!
zepdrix
  • zepdrix
When we started this problem, we weren't dividing fractions. Only the top one was a fraction. Maybe that's why there was a little confusion :)
anonymous
  • anonymous
i think so...this is the only one that i've been confused over so far... thank you:)
anonymous
  • anonymous
i got the right answer. thanks a billion
zepdrix
  • zepdrix
yay!

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