Please prove this...
Given
$\displaystyle x = cy+bz, y = az+cx, z = bx+ay$ ; show that $\displaystyle \frac{x^2}{1-a^2} = \frac{y^2}{1-b^2} = \frac{z^2}{1-c^2}$
~Kalyan
$\displaystyle y = az+cx \Rightarrow y = az+c(cy+bz) = az +cbz + c^2y \Rightarrow y(1 - c^2) = (a+bc)z$-----------(1)
$\displaystyle z = bx+ay = b(cy+bz) + ay = (bc+a)y+ zb^2 \Rightarrow z(1-b^2) = (a+bc)y$----------------------(2)
Divide (1) by (2), to get $\displaystyle \frac{y(1-c^2)}{z(1- b^2)} = \frac{z}{y} \Rightarrow \frac{y^2}{1- b^2} = \frac{z^2}{1 - c^2}$
You can similarly prove the other ratio by eliminating y(or z) in both the equations.