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Mathematics 13 Online
OpenStudy (lindsey11x16):

cos^2[(pi/2)-x] divided by cosx how does this match with sinxtanx

OpenStudy (lindsey11x16):

\[\cos ^2[(\Pi/2)-x]/cosx\]

OpenStudy (raden):

hint : cos(pi/2-x) = sinx

OpenStudy (lindsey11x16):

yeah i know but i'm confused cuz theres a ^2 :/

OpenStudy (lindsey11x16):

does it become sin^2 ?

OpenStudy (raden):

sure, for the numerator be sin^2 x

OpenStudy (lindsey11x16):

so then i get sin^2x / cosx

OpenStudy (raden):

yeah, = sinx * sinx/cosx, right ?

OpenStudy (lindsey11x16):

yes

OpenStudy (lindsey11x16):

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