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Mathematics 22 Online
OpenStudy (anikate):

((cosx*cotx)/(1-sinx)) -1 = cscx verify identity

OpenStudy (anikate):

\[\frac{ cosxcotx }{ 1-sinx } -1 = cscx\]

OpenStudy (anikate):

verify identity

OpenStudy (shamim):

Cosx*cotx=cosx*cosx/sinx=cos^2x/sinx

OpenStudy (anikate):

how?

OpenStudy (anikate):

@shamim

Vocaloid (vocaloid):

cot(x) = cos(x)/sin(x), right? so cos(x)*cot(x) = cos(x)*cos(x)/sin(x) = cos^2x/sinx

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