If cos ^{-1} x+cos ^{-1} y+cos ^{-1} z=pi and | vedclass

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(D) 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

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$2$

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(D) Given,$\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi$ $\Rightarrow \cos ^{-1} x+\cos ^{-1} y=\pi-\cos ^{-1} z$ $\Rightarrow \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}(-z)$ $\Rightarrow \cos ^{-1}\left[x y-\sqrt{1-x^2} \sqrt{1-y^2}\right]=\cos ^{-1}(-z)$ $\Rightarrow x y-\sqrt{(1-x^2)(1-y^2)}=-z$ $\Rightarrow x y+z=\sqrt{(1-x^2)(1-y^2)}$ Squaring both sides: $(x y+z)^2 = (1-x^2)(1-y^2)$ $x^2 y^2+2 x y z+z^2=1-y^2-x^2+x^2 y^2$ $x^2+y^2+z^2+2 x y z=1$ Comparing this with the given equation $x^2+y^2+z^2+k x y z=1$,we get $k=2$.

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

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