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Question Number 27503 by Rasheed.Sindhi last updated on 07/Jan/18

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

Answered by mrW1 last updated on 07/Jan/18

$$-x+cy+bz=0..\left(i\right)$$ $$cx-y+az=0..\left(ii\right)$$ $$bx+ay-z=0..\left(iii\right)$$ $$$$ $$\left(i\right)\times c+\left(ii\right):$$ $$(c^2-1)y+(a+bc)z=0$$ $$(1-c^2)y=(a+bc)z$$ $$(1-c^2)y^2=(a+bc)yz..\left(iv\right)$$ $$$$ $$\left(i\right)\times b+\left(iii\right):$$ $$(a+bc)y+(b^2-1)z=0$$ $$(1-b^2)z=(a+bc)y$$ $$(1-b^2)z^2=(a+bc)yz..\left(v\right)$$ $$$$ $$\Rightarrow (1-c^2)y^2=(1-b^2)z^2$$ $$\Rightarrow \frac{y^2}{1-b^2}=\frac{z^2}{1-c^2}$$ $$similarly$$ $$\Rightarrow \frac{x^2}{1-a^2}=\frac{y^2}{1-b^2}=\frac{z^2}{1-c^2}$$

Commented byRasheed.Sindhi last updated on 07/Jan/18

$$\mathcal{WELDONE}\mathcal{S}i\mathcal{R}!$$

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