How do you verify $sinx+cosx=cosx/(1-tanx)+sinx/(1-cotx)$ ?

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Answer 1 · 2016-11-12T10:50:29

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$sinx+cosx=cosx/(1-tanx)+sinx/(1-cotx)$

Right Side: $=cosx/(1-tanx)+sinx/(1-cotx)$

$=cosx/(1-sinx/cosx)+sinx/(1-cosx/sinx)$

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

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

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

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

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

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

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

$=cosx+sinx$

$=sinx+cosx$

$:.=$ Left Side

How do you verify sinx+cosx=cosx/(1-tanx)+sinx/(1-cotx)sinx+cosx=cosx/(1-tanx)+sinx/(1-cotx)?