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Question

sin1x+sin1y+sin1z=3π2, then the value of x100+y100+z1009x101+y101+z101=


A

0

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B

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C

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D

13

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Solution

The correct option is A

0


sin1x+sin1y+sin1z=3π2
We know that sin1xπ2
sin1x=sin1y=sin1z=π2
x=y=z=sinπ2=1
x100+y100+z1009x101+y101+z101=393=0


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