How do you prove: #cosx- (cosx/(1-tanx))= (sinxcosx)/(sinx-cosx)#?

1 Answer
Sidharth · Ahmed Elnaiem · Becca M.
Apr 14, 2015

#(sinx cosx) / (sinx - cosx) = cosx - [(cosx) / ( 1 - tan x)]#

#(sinx cosx) / (sinx - cosx) = cosx - {(cos x ) / [ 1 - ( sinx / cosx)]} #

#(sinx cosx) / (sinx - cosx) = cosx - {(cosx) / [(cosx - sinx) / cosx]}#

#(sinx cosx) / (sinx - cosx) = cosx - {(cosx) [cosx / (cosx - sinx)]}#

#(sinx cosx) / (sinx - cosx) = cosx - [ (cos ^2 x) / (cosx - sinx)]#

#(sinx cosx) / (sinx - cosx) = [cosx( cosx - sinx ) - (cos ^2 x) ] / (cosx - sinx)#

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

#(sinx cosx) / (sinx - cosx) = - (cosx sinx) / ( cosx - sinx )#

#(sinx cosx) / (sinx - cosx) = - (cosx sinx) / [ - ( sinx - cosx)]#

#(sinx cosx) / (sinx - cosx) = (-cosx sinx) / [ - ( sinx - cosx)]#

#(sinx cosx) / (sinx - cosx) = (cosx sinx) / ( sinx - cosx)#

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