Prove that #(tanx+tany)/(tanx-tany)=2cscx#?

Question

3518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-26T16:35:05

#(tanx+tany)/(tanx-tany)=-cos(x-y)/cos(x+y)#

not #2cscx#

#(tanx+tany)/(tanx-tany)#

= #(sinx/cosx+cosy/siny)/(sinx/cosx-cosy/siny)#

= #((sinxsiny+cosxcosy)/(cosxsiny))/((sinxsiny-cosxcosy)/(cosxsiny))#

= #(cosxcosy+sinxsiny)/-(cosxcosy-sinxsiny)#

= #-cos(x-y)/cos(x+y)#

Hence #(tanx+tany)/(tanx-tany)=-cos(x-y)/cos(x+y)#

not #2cscx#