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cos^2[(pi/2)-x] divided by cosx how does this match with sinxtanx
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\[\cos ^2[(\Pi/2)-x]/cosx\]
hint : cos(pi/2-x) = sinx
yeah i know but i'm confused cuz theres a ^2 :/
does it become sin^2 ?
sure, for the numerator be sin^2 x
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so then i get sin^2x / cosx
yeah, = sinx * sinx/cosx, right ?
yes
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