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

Question

Aicharya · Accepted Answer

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

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

Solution