Question

If x = cy + bz, y = az + cx, z = bx + ay, shew that $\frac{x^2}{1-a^2}=\frac{y^2}{1-b^2}=\frac{z^2}{1-c^2}.$

Answer 1

$x-cy-bz=\mathrm{0.....}\left(i\right)cx-y+az=\mathrm{0.....}\left(ii\right)bx+ay-z=\mathrm{0.....}\left(iii\right)$

Applying cross multiplication in $\left(i\right)$ and $\left(ii\right)$

$\frac{x}{b+ac}=\frac{y}{bc+a}=\frac{z}{1-c^2}.........\left(iv\right)$

From $\left(ii\right)$ and $\left(iii\right)$

$\frac{x}{1-a^2}=\frac{y}{ab+c}=\frac{z}{ac+b}..........\left(v\right)$

From $\left(ii\right)$ and $\left(iii\right)$

$\frac{x}{ab+c}=\frac{y}{1-b^2}=\frac{z}{bc+a}...........\left(vi\right)$

Multiplying $\left(v\right)$ and $\left(vi\right)$

$\frac{x}{1-a^2}\times \frac{x}{ab+c}=\frac{y}{ab+c}\times \frac{y}{1-b^2}\frac{x^2}{1-a^2}=\frac{y^2}{1-b^2}......\left(vii\right)$

Multiplying $\left(iv\right)$ and $\left(vi\right)$

$\frac{y}{bc+a}\times \frac{y}{1-b^2}=\frac{z}{1-c^2}\times \frac{z}{bc+a}\frac{y^2}{1-b^2}=\frac{z^2}{1-c^2}......\left(viii\right)$

From $\left(vii\right)$ and $\left(viii\right)$

$\frac{x^2}{1-a^2}=\frac{y^2}{1-b^2}=\frac{z^2}{1-c^2}$

Hence proved.