$\begin{aligned}
& \text { Given that } \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi \\
& \Rightarrow \cos ^{-1}(x)+\cos ^{-1}(y)+\cos ^{-1}(z)=\cos ^{-1}(-1) \\
& \Rightarrow \cos ^{-1}(x)+\cos ^{-1}(y)=\cos ^{-1}(-1)-\cos ^{-1}(z) \\
& \Rightarrow \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{\left.1-y^2\right)}=\cos ^{-1}\{(-1)(z)\}\right. \\
& \Rightarrow x y-\sqrt{\left(1-x^2\right)\left(1-y^2\right)}=-z \\
& \Rightarrow(x y+z)=\sqrt{\left(1-x^2\right)\left(1-y^2\right)}
\end{aligned}$
Squaring both sides we get $x^2+y^2+z^2+2 x y z=1$
Trick : Put
$x=y=z=\frac{1}{2}$
so that
$\cos ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}=\pi$
Obviously (4) holds for these values of $x, y, z$.