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Question

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

A
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B
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Solution

The correct option is A 0
Given,sin1x+sin1y+sin1z=3π2

This will only possible when sin1x=sin1y=sin1z=π2

Since sin1x[π2,π2]

then =sinπ2=1x=y=z=1

Hence desired value is =1+1+1+(91+1+1)=393=0

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