# if arccos x+ arccos y + arccos z= ╥ then prove x^2 +y^2 +z^2+ 2*x*y*z = 1

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Put arccosx=m, arccosy=n and arccos y=p => `m+n+p=pi`

Subtract p both sides =>`m+n=pi-p`

calculate cosine value of both sides:

`cos(m+n)=cos(pi-p)`

use the formula: `cos(m+n)=cos m*cos n-sin m*sin n`

`cos(pi-p)=cos pi*cos p+sin pi*sin p (cos pi=-1, sin pi=0)`

`cos m*cos n-sin m*sin n=-cos p`

cos m=cos(arccos x)=x

cos n=cos(arccos y)=y

cos p=cos(arccosz)=z

`sin m=sin(arccos x)=sqrt(1-x^2)`

`` `sin n=sqrt(1-y^2)`

`cos m*cos n-sin m*sin n=-cosp ` <=> `xy-sqrt((1-x^2)(1-y^2))=-z`

`sqrt((1-x^2)(1-y^2))=z+xy`

Raise to square both sides to remove the square root:

`(1-x^2)(1-y^2)=(z+xy)^2`

Remove the brackets:

`1-x^2-y^2+x^2*y^2=z^2+2xyz+x^2*y^2`

Reduce the same quantity`x^2*y^2` both sides

`1-x^2-y^2=z^2+2xyz`

Add `x^2+y^2 ` both sides

`x^2+y^2+z^2+2xyz=1`

**Answer: `x^2+y^2+z^2+2xyz=1` proves the identity `arccos x+arccos y+arccos z=pi` .**