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help with secx/sinx-sinx/cosx!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
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you wanna simplify it? note that sec x = 1/cos x so it becomes \(\frac{1}{sin x cos x} - \frac{sin x}{cos x} \) if you continue to put both fractions into one , you'll notice something beautiful.
well i got ((1/cosx)/sinx)-sinx/cos x and IDK what what to do after that unless i change cosx to 1/secx then can i cancel out the two secx?
The sec x cannot be "cancelled". from above, \[\frac{ 1 }{ \sin x \cos x }-\frac{ \sin x }{ \cos x }=\frac{ 1-\sin x(\sin x) }{ \sin x \cos x}\] and \(cos^2 x+sin^2 x =1 \) :) I think this will help.

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but if you subtract sin from 1 how did you get 1-sinx(sinx)? wouldn't it just be 1-sinx?
hmm...I didn't subtract sin x.. \[\frac{ 1 }{ a }-\frac{ 1 }{ b }=\frac{ b-a }{ ab }\] I think this should clear things up.
but you have two things in the first one cuz it's1/sinxcosx not just 1/sinx
yup! :) so I multiplied sin x to \( \frac {sin x}{cos x}\) to have the same denominator.
oh okay then you seperated it into 1-sinx(sinx) because when you multiply it you get 1-sin^2x
but now what? cuz i know it has to equal cotx but i don't see how
well, since from \(cos^2x+sin^2x=1\) we know that \(cos^2 x=1-sin^2 x\) so it simplifies to \(\frac{cos^2}{sin x cos x}\)=cot x
but there's a cosx in the denominator so how does that work if cot x =cosx/sinx?
well, \( \frac{cos^2 x}{ sinxcosx}\)= \(\frac{cos x cosx}{sin x cos x}\) and we can cancel the cos x from the numerator and denominator.
oh okay!!!! weLl thanx
You're welcome :)

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