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Question

If $$cos^{-1}x+cos^{-1}y+cos^{-1}z=\pi$$, then 


A
x2+y2+z2+xyz=0
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B
x2+y2+z2+2xyz=0
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C
x2+y2+z2+xyz=1
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D
x2+y2+z2+2xyz=1
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Solution

The correct option is C $$x^{2}+y^{2}+z^{2}+2xyz=1$$
$$cos^{-1}(x)+ cos^{-1}(y)=\pi-cos^{-1}(z)$$
$$cos^{-1}(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}})=\pi-cos^{-1}(z)$$
Taking cos on both the sides, we get
$$xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}=-z$$
$$xy+z=\sqrt{1-x^{2}}\sqrt{1-y^{2}}$$
Squaring on both the sides, we get
$$x^{2}y^{2}+z^{2}+2xyz=1-y^{2}-x^{2}+x^{2}y^{2}$$
$$x^{2}+y^{2}+z^{2}+2xyz=1$$

Mathematics

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