Question

tanx+tany+tanz=tanx×tany×tanz prove

Answer 1

2 answers · Mathematics 

 Best Answer

tan(x + y + z) 

= [ tan(x+y) + tan z ] /[1 - tan(x+y)tan(z)] 

= [ {tan(x) + tan(y)}/(1 - tan(x)tan(y)) + tan z ] /[1 - tan(z){tan x + tan y) /(1 - tan x tan y) ] 

= [ {tan(x) + tan(y) + tan z - tan(x)tan(y) tan(z) ] /[1 - tan(x) tan(y) - tan(z)tan(x) - tan(y)tan(z)]

LHS = tan(x - y) + tan( y - z) = (tanx - tany) / (a million + tanx tany) + (tany - tanz) / (a million + tany tanz) = [(tanx - tany) (a million + tany tanz)... show

Answer 2

tanx+tany+tanz=tanx×tany×tanz

Proved below.

Step-by-step explanation:

Given:

Let x+y+z=π

  ⇒x+y=π−z

Now taking tan on both side, we get

tan(x+y)=tan(π−x)=−tanz  

tan(x+y)=−tanz

As we know that $[tan(x+y)=\frac{tanx+tany}{1-tanxtany}]$, therefore

tan(x+y)=−tanz

$\frac{tanx+tany}{1-tanxtany}=-tanz$

tanx+tany=-tanz(1-tanxtany)

tanx+tany=-tanz+tanxtanytanz

$tanx+tany+tanz=tanx\times tany\times tanz$

Hence proved.