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Mathematics 19 Online
OpenStudy (anonymous):

[(1+cosx)/(sinx)]+[(sinx)/(1+cosx)]=2cscx

OpenStudy (anonymous):

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OpenStudy (anonymous):

((1+cosx)^2+(sinx)^2)/((1+cosx)*sinx) =(1+2cosx +cos^2x +sin^2x)/((1+cosx)*sinx) =(1+2cosx + 1)/((1+cosx)*sinx) =(2+2cosx)/((1+cosx)*sinx) =2(1+cosx)/((1+cosx)*sinx) =2/sinx =2cosecx

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