# Prove that E=tgx tg2x tg 3x+tg x+tg2x-tg3x do not depend of x?

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You need to re-arrange the terms, such that:

`E = tan x + tan 2x - tan 3x + tan x*tan 2x*tan 3x`

You need to factor out `tan 3x` such that:

`E = tan x + tan 2x - tan 3x(1 - tan x*tan 2x)`

You need to replace `x + 2x` for `3x` such that:

`E = tan x + tan 2x - tan (x + 2x)(1 - tan x*tan 2x)`

You should use the following trigonometric identity, such that:

`tan(a + b) = (tan a + tan b)/(1 - tan a*tan b)`

Reasoning by analogy, yields:

`tan (x + 2x) = (tan x + tan 2x)/(1 - tan x*tan 2x)`

Replacing `(tan x + tan 2x)/(1 - tan x*tan 2x)` for `tan (x + 2x) ` yields:

`E = tan x + tan 2x - (tan x + tan 2x)/(1 - tan x*tan 2x)*(1 - tan x*tan 2x)`

Reducing duplicate factors yields:

`E = tan x + tan 2x - (tan x + tan 2x)`

`E = 0`

**Hence, evaluating the expression E, yields that E does not depend on x.**

**Sources:**

We have given

tgx tg2x tg 3x+tg x+tg2x-tg3x

`E=tan(x)tan(2x)tan(3x)+tan(x)+tan(2x)-tan(3x)`

`=tan(x)tan(2x)tan(3x)-tan(3x)+tan(x)+tan(2x)`

`=tan(3x)(tan(x)tan(2x)-1)+tan(x)+tan(2x)`

`=(tan(x)tan(2x)-1)(tan(3x)+(tan(x)+tan(2x))/(tan(x)tan(2x)-1))`

`=(tan(x)tan(2x)-1)(tan(3x)-(tan(x)+tan(2x))/(1-tan(x)tan(2x)))`

`=(tan(x)tan(2x)-1)(tan(3x)-tan(3x))`

`=(tan(x)tan(2x)-1)(0)`

`=0`

`E=0`

E=0 ,this mean E is independent of x

E =0 does not depend on value of x.

**Sources:**