If #cotx+coty+cotz=0#, prove that #(tanx+tany+tanz)^2=tan^2x+tan^2y+tan^2z#?

Question

2127 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T07:20:33

Please see below.

As #cotx+coty+cotz=0#, we have

#1/tanx+1/tany+1/tanz=0#

i.e. #(tanxtany+tanytanz+tanztanx)/(tanxtanytanz)=0#

or #tanxtany+tanytanz+tanztanx=0#

#(sumsinx/cosx)^2=(tanx+tany+tanz)^2#

and #sum(sinx/cosx)^2=tan^2x+tan^2y+tan^2z#

#:.(tanx+tany+tanz)^2#

#=tan^2x+tan^2y+tan^2z+2tanxtany+2tanytanz+2tanztanx#

#=tan^2x+tan^2y+tan^2z+2xx0#

#=tan^2x+tan^2y+tan^2z#