Question

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

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

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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