If \(\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi\) and \(x^2+y^2+z^2+k x y z=1\), then k is

\(\begin{aligned} & \text { Given, } \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi \\ & \Rightarrow \cos ^{-1} x+\cos ^{-1} y=\pi-\cos ^{-1} z \\ & \Rightarrow \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}(-z) \\ & \Rightarrow \cos ^{-1}\left[x y-\left(\sqrt{1-x^2}\right)\left(\sqrt{1-y^2}\right)\right]=\cos ^{-1}-z \\ & \Rightarrow x y-\sqrt{\left(1-x^2\right)\left(1-y^2\right)}=-z \\ & \Rightarrow x y+z=\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\end{aligned}\)
\(\begin{array}{ll}
& \text { Squaring on both sides } \\
& x^2 y^2+2 x y z+z^2=\left(1-x^2\right)\left(1-y^2\right) \\
& x^2 y^2+2 x y z+\mathrm{z}^2=1-y^2-x^2+x^2 y^2 \\
\therefore \quad & x^2+y^2+\mathrm{z}^2+2 x y \mathrm{z}=1 \\
\quad & \text { But } x^2+y^2+\mathrm{z}^2+\mathrm{kxyz}=1 \\
\therefore \quad \mathrm{k}=2
\end{array}\)