Given #x=bz+cy;" " y=az+cx and z=ay+bx# how will you prove #x^2/(1-a^2)=y^2/(1-b^2)=z^2/(1-c^2)#?

1 Answer
Dec 19, 2016

Answer:

See below.

Explanation:

#{(x=bz + c y),(y = az + cx),(z = ay + bx):}#

solving for couples we obtain

(1-1) #{(x = ((b + a c) z)/(c^2-1)), (y = ((a + b c) z)/(c^2-1)):}#

(1-2) #{(y = ((a b + c) x)/(a^2-1)), (z = ((b + a c) x)/(a^2-1)):}#

(1-3) #{(z = ((a + b c) y)/(b^2-1)), (x = ((a b + c) y)/(b^2-1)):}#

From (1-1)

#z/(c^2-1)=x/(b+ac) -> z^2/(c^2-1)= (x z)/(b+ac)#

From (1-2)

#x/(a^2-1)=z/(b+ac)->x^2/(c^2-1) = (x z)/(b+ac)#

and also

#z/(c^2-1)=y/(a+bc)->z^2/(c^2-1)=(yz)/(a+bc)#

#y/(b^2-1)=z/(a+bc)->y^2/(b^2-1) = (yz)/(a+bc)#

then follows

#x^2/(a^2-1)=y^2/(b^2-1)=z^2/(c^2-1)#