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Precalculus 13 Online
OpenStudy (anonymous):

(1+sinx)/cosx=cosx/(1-sinx)

OpenStudy (anonymous):

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OpenStudy (anonymous):

\[1-\sin ^{2}x=\cos ^{2}x\]

OpenStudy (anonymous):

you cant do that

OpenStudy (anonymous):

lol actually what's the question?

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