Question #5abd9

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Answer 1 · 2017-11-14T02:26:34

The right side of the equation was missing a minus sign between the $cotx$ and $tanx$ .

$(1-2(sinx)^2)/((sinx)(cosx))$

$=1/((sinx)(cosx))-(2(sinx)^2)/((sinx)(cosx))$

$=((sinx)^2+(cosx)^2)/((sinx)(cosx))-2tanx$

$=((sinx)^2/((sinx)(cosx)))+((cosx)^2/((sinx)(cosx)))-2tanx$

$=((sinx)/(cosx))+((cosx)/(sinx))-2tanx$

$=tanx+cotx-2tanx$

$=cotx-tanx$

Answer 2 · 2017-11-14T06:21:33

$LHS=(1-2sin^2x)/(sinxcosx)$

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

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

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

$=cosx/sinx-sinx/cosx$

$=cotx-tanx$