If #tanx=-12/5#, find #((cosx-1)(cosx+1))/((sinx-1)(sinx+1))#?

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Answer 1 · 2016-06-29T05:41:19

#((cosx-1)(cosx+1))/((sinx-1)(sinx+1))=144/25#

As #tanx=-12/5#, we have

#cosx=1/secx=1/sqrt(sec^2x)=1/sqrt(1+tan^2x)#

= #1/sqrt(1+(-12/5)^2)=1/sqrt(1+144/25)=1/sqrt(169/25)#

= #+-1/(13/5)=+-5/13# and

#sinx=sinx/cosx xxcosx=tanx xxcosx=-12/5xx(+-5/13)=+-12/13#

or #cos^2x=25/169# and #sin^2x=144/169#

Hence #((cosx-1)(cosx+1))/((sinx-1)(sinx+1))#

= #(cos^2x-1)/(sin^2x-1)=(25/169-1)/(144/169-1)#

= #(-144/169)/(-25/169)=-144/169xx-169/25=144/25#

Answer 2 · 2016-06-29T13:57:02

#144/25#

#((cosx-1)(cosx+1))/((sinx-1)(sinx+1))#

#=-((1-cosx)(1+cosx))/-((1-sinx)(1+sinx))#
#=(1-cos^2x)/(1-sin^2x)#

#=sin^2x/cos^2x=tan^2x=(-12/5)^2=144/25#