Mathematics 76 Online 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):

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