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

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

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

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

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

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

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