Question

If three real numbers $a,b,c$ none of which is zero are related by: $a^2=b^2+c^2-2bc\sqrt{1-a^2},b^2=c^2+a^2-2ca\sqrt{1-b^2},c^2=a^2+b^2-2ab\sqrt{1-c^2}$, then prove that $a=c\sqrt{1-b^2}+b\sqrt{1-c^2}$.

Answer 1

$a^2=b^2+c^2-2bc\sqrt{1-a^2}$........(1)

$b^2=c^2+a^2-2ca\sqrt{1-b^2}$........(2)

$c^2=a^2+b^2-2ab\sqrt{1-c^2}$........(3)

$\left(1\right)+\left(2\right)+\left(3\right),$

$\Rightarrow a^2+b^2+c^2=2(a^2+b^2+c^2)$$-2(ab\sqrt{1-c^2}+bc\sqrt{1-a^2}+ca\sqrt{1-b^2})$

$\Rightarrow a^2+b^2+c^2$ $=2(ab\sqrt{1-c^2}+bc\sqrt{1-a^2}+ca\sqrt{1-b^2})$

$\Rightarrow \frac{a^2+b^2+c^2}{2abc}=\frac{\sqrt{1-c^2}}{c}+\frac{\sqrt{1-a^2}}{a}+$ $\frac{\sqrt{1-b^2}}{b}$

$\Rightarrow \frac{a^2+b^2+c^2}{2abc}=\frac{\sqrt{1-a^2}}{a}+\frac{\sqrt{1-b^2}}{b}+\frac{\sqrt{1-c^2}}{c}$ ........(4)

From equation (1)

$a^2=b^2+c^2-2bc\sqrt{1-a^2}$

$\Rightarrow b^2+c^2=a^2+2bc\sqrt{1-a^2}$

L.H.S $=\frac{a^2+(b^2+c^2)}{2abc}=\frac{a^2+a^2+2bc\sqrt{1-a^2}}{2abc}$

R.H.S $=\frac{\sqrt{1-a^2}}{a}+\frac{\sqrt{1-b^2}}{b}+\frac{\sqrt{1-c^2}}{c}$

$\Rightarrow \frac{a^2}{abc}+\frac{bc\sqrt{1-a^2}}{abc}=\frac{\sqrt{1-a^2}}{a}+\frac{\sqrt{1-b^2}}{b}+\frac{\sqrt{1-c^2}}{c}$

$\Rightarrow \frac{a}{bc}=\frac{\sqrt{1-b^2}}{b}+\frac{\sqrt{1-c^2}}{c}$

$\Rightarrow a=c\sqrt{1-b^2}+\sqrt{1-c^2}$