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what is (cosx/sinx)/sinxcosx simplified

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\[\frac{ \cos(x) }{ \sin(x) } \times \frac{ \sin(x)\cos(x) }{ 1 }\]
\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x}\]Woops I dunno if you're looking at it correctly abbot :O Only the top one is a fraction.
\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x} \qquad = \qquad \left(\frac{\cos x}{\sin x}\right)\frac{1}{\sin x \cos x}\]

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Other answers:

ya but what about multiplying by the recipricol.... cause i am seriously confused.
\[\large \left(\frac{\cancel{\cos x}}{\sin x}\right)\frac{1}{\sin x \cancel{\cos x}}\]
Ok here's the thing with the reciprocal :) lemme explain.
ok sounds good
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.
aha!!!!! now it's making sense!!!
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 :)
i think so...this is the only one that i've been confused over so far... thank you:)
i got the right answer. thanks a billion

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