How do you simplify the expression ##(sinx-cosx)/(sinxcosx)##?

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##frac{sinx-cosx}{sinx cosx} = secx – cscx##

##frac{sinx-cosx}{sinx cosx} = frac{sinx}{sinx cosx} – frac{cosx}{sinx cosx}##

##= frac{1}{cosx} – frac{1}{sinx}##

##= secx – cscx##

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