A) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+xyz=0\]
B) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=0\]
C) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+xyz=1\]
D) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1\]
Correct Answer: D
Solution :
[d] Given that \[{{\cos }^{-1}}(x)+co{{s}^{-1}}(y)+co{{s}^{-1}}(z)=\pi \] \[\Rightarrow {{\cos }^{-1}}(x)+co{{s}^{-1}}(y)+co{{s}^{-1}}(z)={{\cos }^{-1}}(-1)\] \[\Rightarrow {{\cos }^{-1}}(x)+co{{s}^{-1}}(y)=\pi -co{{s}^{-1}}(z)\] \[\Rightarrow {{\cos }^{-1}}(xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}})=co{{s}^{-1}}(-z)\] \[\Rightarrow xy-\sqrt{(1-{{x}^{2}})(1-{{y}^{2}})}=-z\] \[\Rightarrow (xy+z)=\sqrt{(1-{{x}^{2}})(1-{{y}^{2}})}\] Squaring both sides, we get \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1.\]
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