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verify that (sinxcosx+cosx)/(sinx+sin^2x)=cotx
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\[\Large \frac{\sin x \cos x + \cos x}{\sin x + \sin^2x}\quad=\]Factoring a cosine out of each top term, and a sine out of each bottom term gives us, \[\Large \frac{\cos x(\sin x+1)}{\sin x(1+\sin x)}\quad=\]
Do you see the cancellation we can make from this spot?
(sinxcosx+cosx)/(sinx+sin^2x) =cosx(sinx+1)/sinx(1+sinx) =cosx/sinx =cotx
yes! we can cancel out the (sinx-1) so that we get \[\frac{ \cos x }{ \sin x } \]
yes good :)
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