Prove that $(cotx-tanx)/(sinx+cosx)=cscx-secx$ ?

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Answer 1 · 2016-05-03T09:46:18

Please see below.

$(cotx-tanx)/(sinx+cosx)$

= $(cosx/sinx-sinx/cosx)/(sinx+cosx)$

= $((cos^2x-sin^2x)/(sinxcosx))/(sinx+cosx)$

= $((cosx-sinx)(cosx+sinx))/(sinxcosx)xx1/(sinx+cosx)$

= $((cosx-sinx)cancel((cosx+sinx)))/(sinxcosx)xx1/cancel((sinx+cosx))$

= $((cosx-sinx))/(sinxcosx)$

= $cosx/(sinxcosx)-sinx/(sinxcosx)$

= $1/sinx-1/cosx$

= $cscx-secx$

Prove that (cotx-tanx)/(sinx+cosx)=cscx-secx(cotx-tanx)/(sinx+cosx)=cscx-secx?