If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$ If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\pi-\arcsin z$
$x\sqrt{1-y^2}+y\sqrt{1-x^2}=z$
Similarly,$y\sqrt{1-z^2}+z\sqrt{1-y^2}=x$
Similarly,$x\sqrt{1-z^2}+z\sqrt{1-x^2}=y$
Adding the three equations,we get
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=\frac{x+y+z}{2}$
I am stuck here,please help me.Thanks.
 A: Let $\sin^{-1}(x) = A\Rightarrow x=\sin A$ and $\sin^{-1}(y)=B\Rightarrow y=\sin (B)$
and $\sin^{-1}(z)=C\Rightarrow z=\sin C$
So above we have given $A+B+C = \pi$
Now we have to prove $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$
So for $\bf{L.H.S}$
$$\sin 2A+\sin 2B+\sin 2C = 2\sin (A+B)\cdot \sin (A-B)+2\sin C\cos C$$
$$=2\sin C\cdot \sin (A-B)+2\sin C\cos C$$
$$=2\sin C\left[\cos(A-B)+\cos(\pi-(A+B))\right]$$
$$=4\sin C\cdot \left[\cos(A-B)-\cos(A+B)\right]$$
$$=4\sin A\sin B\sin C$$
A: Hint:
Write
$$x = \sin X \qquad y = \sin Y \qquad z = \sin Z$$
Where $X + Y + Z = \pi$ (meaning that those are the angles of a triangle).
A: All we need to facilitate analysis is the double angle formula 
$$\sin(2a)=2\sin(a)\,\cos(a) \tag 1$$
the Prosthaphareis Identity
$$\cos (a-b)-\cos(a+b)=2\sin (a)\,\sin (b) \tag 2$$
along with the Reverse Prosthaphareis Identity
$$\sin (a)+\sin (b)=2\sin\left(\frac{a+b}{2}\right)\,\cos\left(\frac{a-b}{2}\right) \tag 3$$
Then, letting $x=\sin \alpha$, $y=\sin \beta$, and $z=\sin \gamma$ gives
$$\alpha +\beta +\gamma = \pi \tag 4$$
We wish now to transform the function $F(\alpha,\beta,\gamma)$ as given by
$$\begin{align}
F(\alpha,\beta,\gamma)&=\frac12 \sin (2\alpha)+\frac12 \sin (2\beta)+\frac12 \sin (2\gamma) \tag 5\\\\
&=x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}
\end{align}$$
First, use $(1)$ with $a=2\gamma$, $(3)$ with $a=2\alpha$ and $b=2\beta$, and $(4)$ to write $(5)$ as
$$\begin{align}
F(\alpha,\beta,\gamma)&=\sin(\alpha+\beta)\,\cos(\alpha-\beta)+\sin(\gamma)\,\cos (\gamma) \\\\
&=\sin(\pi-\gamma)\,\cos(\alpha-\beta)+\sin(\gamma)\,\cos (\pi-\alpha-\beta)
\end{align}$$
Next, substituting $(4)$ into $(6)$ and recalling that $\sin (\pi -a)=\sin(a)$ and $\cos (\pi-a)=-\cos(a)$ reveals
$$F(\alpha,\beta,\gamma)=\sin(\gamma)\left(\cos(\alpha-\beta)-\cos (\alpha+\beta)\right) \tag 7$$
Finally, using $(2)$ in $(7)$ yields
$$\begin{align}
F(\alpha,\beta,\gamma)&=\sin(\gamma)\left(2\sin(\alpha)\,\sin(\beta)\right)\\\
&=2\sin(\alpha)\,\sin(\beta)\,\sin)\gamma)\\\\
&=2xyz
\end{align}$$
Therefore, we obtain the identity
$$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$$
as was to be shown!
