<s>[INST] 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.
 [/INST] 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 $\bf{L.H.S}$
$$\sin 2A+\sin 2B+\sin 2C = 2\sin (A+B)\cdot \sin (A-B)+2\sin C\cos C$$
$$=2\sin C\cdot \sin (A-B)+2\sin C\cos C$$
$$=2\sin C\left[\cos(A-B)+\cos(\pi-(A+B))\right]$$
$$=4\sin C\cdot \left[\cos(A-B)-\cos(A+B)\right]$$
$$=4\sin A\sin B\sin C$$
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