Question
If cos−1x+cos−1y+cos−1z=3π, then xy+yz+zx=
Answer: Option C
:
C
Given cos−1x+cos−1y+cos−1z=3π
∵0≤cos−1x≤π
∴0≤cos−1y≤π and 0≤cos−1z≤π
Here cos−1x=cos−1y=cos−1z=π
⇒x=y=z=cosπ=−1
∴xy+yz+zx=(−1)(−1)+(−1)(−1)+(−1)(−1)
=1+1+1=3.
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:
C
Given cos−1x+cos−1y+cos−1z=3π
∵0≤cos−1x≤π
∴0≤cos−1y≤π and 0≤cos−1z≤π
Here cos−1x=cos−1y=cos−1z=π
⇒x=y=z=cosπ=−1
∴xy+yz+zx=(−1)(−1)+(−1)(−1)+(−1)(−1)
=1+1+1=3.
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