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

1 Answer
Feb 17, 2016

Please see below for the proof. Feel free to ask questions if you have any.

Explanation:

1) replace tanx with #sinx/cosx#

#cosx-(cosx/(1-tanx))#
= #cosx - (cosx/(1-(sinx/cosx)))#

2) equalise the denominator in the paranthesis

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

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

3) equalise the denominator once more
=#cosx(cosx-sinx) /(cosx-sinx)- cos^2x/(cosx-sinx)#
=#(cos^2x-cosx*sinx-cos^2x)/(cosx-sinx)#
=#(-(cosx*sinx))/(cosx-sinx)#

4) put the denominator into -1 paranthesis
=#-(cosx*sinx)/(-(-cosx+sinx))#

5) - divided by - yields +. And change the places of cosx and sinx in the denominator

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

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