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OpenStudy (anonymous):

y=(cosx)/(1-sinx) use product rule. I got (1)/(1+sinx) but I need verification

OpenStudy (anonymous):

quotent rule

OpenStudy (anonymous):

/?

OpenStudy (anonymous):

yeah I ment quotient rule, my bad

OpenStudy (anonymous):

its wrong wait

OpenStudy (anonymous):

(1-sinx)(-sinx)-(cosx)(-cosx)/(1-sinx)^2

OpenStudy (anonymous):

you always have to square the denominator don't forget

OpenStudy (anonymous):

yeah but we have to simplify using the trig identities

OpenStudy (anonymous):

oh

OpenStudy (zarkon):

\[\frac{1}{1-\sin(x)}\]

OpenStudy (anonymous):

yeah so I'm still uncertain

OpenStudy (anonymous):

how did you get that @zarkon?

OpenStudy (zarkon):

just simplify alvinstifla's solution

OpenStudy (anonymous):

how exactly? please show me steps.

OpenStudy (zarkon):

\[\frac{(1-\sin(x))(-\sin(x))-(\cos(x))(-\cos(x))}{(1-\sin(x))^2}\] \[=\frac{-\sin(x)+\sin^2(x)+\cos^2(x)}{(1-\sin(x))^2}\] \[=\frac{-\sin(x)+1}{(1-\sin(x))^2}\] \[=\frac{1-\sin(x)}{(1-\sin(x))^2}\] \[=\frac{1}{1-\sin(x)}\]

OpenStudy (anonymous):

so how did you end up having 1-sinx as the denominator?

OpenStudy (zarkon):

\[\frac{a}{a^2}=\frac{1}{a}\]

OpenStudy (anonymous):

oh okay so you had (1-sinx)/(1-sinx)(1-sinx) and you cancelled (1-sinx) out to leave you with (1-sinx( right

OpenStudy (zarkon):

yes..you can say it that way

OpenStudy (anonymous):

sorry I had some brain fart or something. For some reason I was changing the signs. Thank you lots Zarkon! do you think you can help me on one more problem?

OpenStudy (anonymous):

OpenStudy (anonymous):

Please go through it.

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