Q.If \(x = cy + bz\), \(y = az + cx\), and \(z = bx + ay\), then prove that \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} \]

Given: \(x = cy + bz \) — (i) \(y = az + cx \) — (ii) \(z = bx + ay \) — (iii) From equation (i): \[ x = cy + bz \] Substituting the value of \(z\) from equation (iii): \[ x = cy + b(bx + ay) \] \[ x = cy + b^2x + aby \] Multiplying both sides by \(x\): \[ x^2 = cxy + b^2x^2 + abxy \quad \text{— (iv)} \] From equation (ii): \[ y = az + cx \] Substituting the value of \(z\) from equation (iii): \[ y = a(bx + ay) + cx \] \[ y = abx + a^2y + cx \] Multiplying both sides by \(y\): \[ y^2 = abxy + a^2y^2 + cxy \quad \text{— (v)} \] Now subtract equation (v) from equation (iv): \[ x^2 - y^2 = b^2x^2 - a^2y^2 \] \[ x^2 - b^2x^2 = y^2 - a^2y^2 \] \[ x^2(1 - b^2) = y^2(1 - a^2) \] \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} \] ( proved)
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